Just use s = ut - 1/2 g t^2
Here s = 4.8 m, u = 1.22 m/s and of course g = 9.8 m/s^2
Plug and you will get a quadratic equation. Right from this you can get the value of t.
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)
Mark off the x-axis in seconds. Mark off the y-axis in meters per second. The graph is a straight line, starting at the origin, sloping down with a slope of -9.8 meters per second per second. It never gets above the x-axis, and 'y' is negative everywhere on the graph. That's because the velocity is always negative (speed directed downward).
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
when flight moves the velocity of air above and below the plane is different ..this gives it an upward thrust (according to bernauli's theorem).
upward
1 sec : position = 10.1 metres above your hand, velocity = 5.2 ms^-1.40 sec : position = 7240 metres below your hand, velocity = 377 ms^-1 downwards.
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
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)
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)
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.
Height reached = 3.7 metres.The mass of the ball is not really relevant.
31 m/s
At the maximum height the ball will be completely stopped from moving upward or downward; thus the speed of the ball would be 0 mph. The ball is only stopped for a split second and then it begins moving downward, then increasing at 9.81m/s^2 until it reaches maximum velocity.
Mark off the x-axis in seconds. Mark off the y-axis in meters per second. The graph is a straight line, starting at the origin, sloping down with a slope of -9.8 meters per second per second. It never gets above the x-axis, and 'y' is negative everywhere on the graph. That's because the velocity is always negative (speed directed downward).
42
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?