A ball is thrown straight up. If the launch velocity is 15 m/s, at what VELOCITY will the ball return to the thrower’s hand?
same speed as it left with, but downward
To find the velocity at which the ball will return to the thrower's hand, we need to consider the motion of the ball. Since the ball is thrown straight up, it will experience free fall due to the force of gravity acting on it.
We can use the equations of motion to find the velocity at which the ball will return to the thrower's hand. The equation for the final velocity (v) in an upward motion can be given by:
v = u - gt
Where:
v is the final velocity,
u is the initial velocity (launch velocity),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken.
In this case, the ball will reach its highest point when its velocity becomes 0. At this point, it will start to fall back down towards the thrower's hand. The time taken to reach the highest point is the same as the time taken for the ball to return to the thrower's hand.
First, let's find the time taken to reach the highest point. We can use the equation:
v = u - gt
0 = 15 - 9.8t
9.8t = 15
t = 15 / 9.8
t ≈ 1.53 seconds
Since it takes 1.53 seconds for the ball to reach its highest point and fall back to the thrower's hand, we can substitute this value back into the equation to find the velocity at this time:
v = u - gt
v = 15 - 9.8 (1.53)
v ≈ 15 - 14.94
v ≈ 0.06 m/s
Therefore, the velocity at which the ball will return to the thrower's hand is approximately 0.06 m/s.