0
An object is thrown vertically upwards from the top of a tower of height 39.2 meters reach the ground in 4 secons Find the velocity with which it is thrown upwards?
Wiki User
∙ 15y ago
Using the equation:
x=vot+(1/2)at2
x: -39.2m (though the object is thrown upwards, its total displacement is just the amount it fell from the tower).
vo: initial velocity
t: 4 s
a: -9.8m/s2 (assuming we're on the earth's surface)
-39.2=vo(4 s)+(1/2) (-9.8m/s2)(4 s)2
((78.4 m - 39.2 m)/ 4 s) = v0
So the initial velocity is 9.8 m/s.
Wiki User
∙ 15y ago
Add your answer:
Earn +20 pts
How do you calculate tak-off velocity of airplane?
peak height = (take off velocity^2)/(2*gravity)
When dropped from this height the kinetic energy when it reaches the ground is equal to if the car was traveling at 100 km h The height from which the 1000 kg car is what?
As long as the mass is the same in both cases, it doesnt enter the calculations.The height (s) from which to drop it so its velocity at impact is 100 kph (27.78 metres per second):Use >s = (v2 - u2) / (2 * a)s = 771.73 / 19.64s = 39.29 metres>u = 0 (initial velocity - metres per second)v = 27.78 (final velocity - metres per second)a = 9.82 (acceleration due to gravity - (m/s)/s)s = ? (distance - in this case , height)
What height will it gain after the escape velocity becomes half?
That will depend not only on the escape velocity, but also - very importantly - on the object's speed.
Calculate distance from a velocity time graph?
The area between the graph and the x-axis is the distance moved. If the velocity is constant the v vs t graph is a straight horizontal line. The shape of the area under the graph is a rectangle. For constant velocity, distance = V * time. Time is the x-axis and velocity is the y-axis. If the object is accelerating, the velocity is increasing at a constant rate. The graph is a line whose slope equals the acceleration. The shape of the graph is a triangle. The area under the graph is ½ * base * height. The base is time, and the height is the velocity. If the initial velocity is 0, the average velocity is final velocity ÷ 2. Distance = average velocity * time. Distance = (final velocity ÷ 2) * time, time is on the x-axis, and velocity is on the y-axis. (final velocity ÷ 2) * time = ½ time * final velocity ...½ base * height = ½ time * final velocity Area under graph = distance moved Most velocity graphs are horizontal lines or sloping lines.
A ball of mass m rolls off a cliff of height h with a horizontal velocity v it reaches the ground with vertical speed of?
If one assumes air resistance to be negligible, then: final velocity = sqrt( g * 2 * h ) where g is 9.8 metres per second per second. The quantities v and m do not matter, because gravitational acceleration does not depend on mass (all objects fall at the same rate) and because the horizontal velocity is independent of the vertical velocity.
A boy kicks a ball vertically upwards with an initial velocity of 12m/s. Calculate the time taken by the ball to reach the maximum hight and the maximum, height reached by the ball?
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
A ball is thrown vertically upwards with velocity of 19.6ms what is the maximum height to which it will reach?
apply the formula: v2 - u2 = 2as. Here v= 0; u = 19.6m/s; a = -g ,find s and that's max. heigth
An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?
An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?
The velocity of a body thrown vertically upward is reduced into half in one secondwhat is the maximum height attained by it?
velocity is found by dividing the distance with time. In a second the height traveled is found by multiplying the velocity by the time taken and then dividing the answer by two.
What height is reached by a 4.0kg ball that is thrown vertically up into the air with an initial velocity of 8.5ms1 if by the time it is height (h) metres above the starting point it has a velocity of?
Height reached = 3.7 metres.The mass of the ball is not really relevant.
What is tje velocity of a ball thrown upward at 16 ft/sec?
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
How do you find the height of something?
Measure upwards from the ground to the top of what ever it is and at the top, mark where it stops then, voiala! You have the height of the thing.
An object is thrown vertically upward when does it reach maximum height?
when the object reaches maximum height, the velocity of the object is 0 m/s.It reaches maximum height when the gravity of the body has slowed its velocity to 0 m/s. If there is no gravity and there is no external force acting on it then it will never reach a maximum height as there wont be a negativeaccelerationdemonstrated by newtons first law.Where there is and you have the objects initial velocity then you can use :v^2 = u^2+2.a.sv = Velocity when it reaches Max. height so v = 0u = Initial Velocity (m/s)a = Retardation/ Negative Acceleration due to gravity, -9.80m/s ^2And then the unknown (s) is the displacement, or height above ground, and if everything else is in the right format it should be in metres.
What is the rock's approximate velocity just prior to hitting the ground?
To answer this question one would need to know the rock's initial height and velocity.
What happens when you release a ball from a height at rest?
It will fall with increasing velocity due to gravity and reach the peak velocity just before hitting the ground.
What do you you need to know to determine how far a projectile travels horizontally?
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
What do you need to know to determine how far a projectile travels horizontally?
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
