a ball is thrown straight up at 50 m/s. What speed will it be when it returns to the thrower's hand?
V = Vo = 50 m/s.
5 seconds
To find the speed of the ball when it returns to the thrower's hand, we need to consider the natural deceleration due to gravity.
When the ball is thrown straight up, its initial velocity is 50 m/s. As it reaches the maximum height, its velocity becomes 0 m/s momentarily before it reverses direction due to gravity. The ball then starts to fall back down towards the thrower's hand.
The acceleration due to gravity is approximately 9.8 m/s², acting downwards. As the ball falls back towards the thrower's hand, it will increase in speed due to this acceleration.
At any point during the ball's descent, we can calculate its speed using the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the ball starts from rest at the maximum height, the initial velocity (u) is 0 m/s. The acceleration (a) is -9.8 m/s² (negative because it acts in the opposite direction to the initial velocity).
To calculate the time it takes for the ball to fall back down, we can use the equation:
s = ut + (1/2)at²
Where:
s = displacement (total distance)
u = initial velocity
a = acceleration
t = time
Since the initial velocity (u) is 0 m/s and the displacement (s) is the same distance the ball was thrown upwards (assuming no air resistance), we can rearrange the equation:
s = (1/2)at²
2s/a = t²
√(2s/a) = t
Now, we can substitute the calculated time (t) back into the first equation to find the final velocity (v) when the ball returns to the thrower's hand:
v = u + at
Let's calculate it step-by-step:
1. Calculate the time it takes for the ball to fall back down:
- Substitute the values into the equation: √(2s/a) = t
- The distance (s) the ball traveled upwards is unknown, so let's assume it's 50 meters (since it was thrown upwards at 50 m/s).
- The acceleration (a) is -9.8 m/s².
- Plug in the values: √(2 * 50 / (-9.8)) = t
- Calculate: √(100 / (-9.8)) = t
- t ≈ √(-10.2) ≈ √(-1) * √(10.2) ≈ i * 3.19 (taking the positive square root)
- t ≈ 3.19 seconds (approximation)
2. Calculate the final velocity (v):
- Substitute the values into the equation: v = u + at
- The initial velocity (u) is 0 m/s (the ball momentarily stops at the maximum height).
- The acceleration (a) is -9.8 m/s² (opposite direction).
- The time (t) is 3.19 seconds.
- Plug in the values: v = 0 + (-9.8) * 3.19
- Calculate: v ≈ -31.26 m/s
Therefore, when the ball returns to the thrower's hand, its speed will be approximately 31.26 m/s in the downward direction (opposite to the initial throw).