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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 high can a ball go?
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
What is the numerical ratio of a velocity to speed of an object?
Velocity is an object's Speed + Direction. For example, a cannon ball fired up at 45o to the ground may have a speed of 1000 ft/sec. However, its forward Velocity is:(1000 ft/sec) x (Sine 45o) = (1000 ft/sec) x (.707)= 707 ft/secSo, the ratio of the horizontal Velocity to the instantaneous Speed of the object is the SINE value of the angle of motion relative to the Horizontal axis.But you have to remember that there is a Vertical component of the object's Speed as well. In this case, the ratio is the COSINE 45o , which happens to be .707 as well. So the ball is going forward at 707 ft/sec, and upward at 707 ft/sec. at the same time.
A 250 gram ball travels at a velocity of 40 ms. It momentum is?
Momentum = (mass) times (velocity) = 0.25 x 40 = 10 kg-m/sec
What was its initail velocity of a ball thrown straight up reaches its highest point 2.6 sec neglect air resistance?
25.48m/s How: At the maximum height the velocity of the ball wil be zero for that instant in time so when t=2.6 v=0 Accleration due to gravity (a) is always equal to -9.8m/s^2 Now you have 3 knowns and 1 unknow find the equation that has these four: a=(v-u)/t substitute the values and you will get 25.48m/s for u
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 high can a ball go?
The maximum height of a thrown ball is dependent on the upward portion of the initial velocity. Air friction will slow it somewhat but gravity will cause it to lose most of its upward velocity. The velocity will decrease by 9.8 m/sec for each second of its upward motion, until it reaches zero. At that point, the ball is pulled back toward Earth.
When a ball is thrown vertically upward at initial velocity of 15 meter per second Find the position and velocity after 1.0 second and 40 second after leaving your hand?
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 velocity of a baseball 4 seconds after it is thrown vertically upward with a speed of 32.1 meter per sec is?
V = V0 + a t V0 = + 32.1 m/sec a = - 9.78 m/sec2 t = 4 sec V = (32.1) + 4 (-9.78) = (32.1) - (39.12) = - 7.02 m/sec (7.02 m/sec downward)
How does gravity affect how high a ball bounces?
Gravity is the force opposing the upward velocity of the ball, which has mass m The ball bounces at velocity v, so kinetic energy = 1/2m x v squared If the ball rises height h, energy to do this is m x g x h These two terms are equal, so solve for h. Using SI units, g = 9.81 meters/sec/sec, m in kg, h in meters, v in meters/sec This ignores air resistance
You throw a ball upward with an initial speed of 20 m sec About 4 seconds later the ball returns and you catch it How fast was the ball traveling downward when you caught it?
20m/sec
What is the numerical ratio of a velocity to speed of an object?
Velocity is an object's Speed + Direction. For example, a cannon ball fired up at 45o to the ground may have a speed of 1000 ft/sec. However, its forward Velocity is:(1000 ft/sec) x (Sine 45o) = (1000 ft/sec) x (.707)= 707 ft/secSo, the ratio of the horizontal Velocity to the instantaneous Speed of the object is the SINE value of the angle of motion relative to the Horizontal axis.But you have to remember that there is a Vertical component of the object's Speed as well. In this case, the ratio is the COSINE 45o , which happens to be .707 as well. So the ball is going forward at 707 ft/sec, and upward at 707 ft/sec. at the same time.
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
If a ball is thrown into the air with a velocity of 38 ft per sec its height in feet after t seconds is given by y equals 38t minus 16t2 Find the velocity when t equals 2?
Substitute the 2 in for t.38(2)-16(2)^276-64=12
A ball is thrown vertically upward. What are its velocity and acceleration when it reaches its maximum altitude What is the acceleration of the ball just before it hits the ground?
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.
What shape does the path of a thrown ball follow?
The ball follows a parabolic path when thrown. In a vacuum (with no air or other forces acting upon it) the gravitational pull of the earth causes the ball to accelerate toward the earth (9.8m/sec
A girl threw a ball upward with an initial velocity of 2000cm/sec and was able to catch it before it reached the ground on its return.C.) how far was the maximum height from starting point?
algebra 2 right? i hated that unit man i forgot everything we learned in that class
