A boy throws a ball straight up into the air so that it leaves his hand at 11 m/s.
1) If the ball has mass 0.15 kg and the boy's arm moved through 1.5 m as he threw the ball, then what average force did he exert on the ball?
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To find the average force exerted by the boy on the ball, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Step 1: Calculate the acceleration of the ball.
Since the ball was thrown vertically upwards, it experiences a downward gravitational force (9.8 m/s^2) acting against its motion. Therefore, the net acceleration will be the difference between the acceleration due to gravity and the deceleration caused by the ball's initial upward velocity.
The deceleration due to the ball's initial upward velocity can be calculated using the formula:
a = (v_f - v_i) / t
where:
a = acceleration
v_f = final velocity (0 m/s, when the ball reaches its maximum height)
v_i = initial velocity (11 m/s, when the ball leaves the boy's hand)
t = time taken to reach the maximum height (which is the same as the time taken to return to the boy's hand)
Since the ball goes up and comes back down, the time taken for a complete cycle will be twice the time taken to reach the maximum height.
Step 2: Calculate the time taken for a complete cycle.
The time taken for the ball to reach the maximum height (t_h) can be calculated using the formula:
t_h = v_i / g
where:
v_i = initial velocity (11 m/s)
g = acceleration due to gravity (9.8 m/s^2)
The time taken for a complete cycle (t_cycle) is then:
t_cycle = 2 * t_h
Step 3: Calculate the acceleration (a) of the ball.
Substituting the values into the formula, we have:
a = (0 - 11) / t_cycle
Step 4: Calculate the average force (F) exerted by the boy on the ball.
Using Newton's second law of motion, we have:
F = m * a
where:
m = mass of the ball (0.15 kg)
a = acceleration of the ball
Substituting the values, we can calculate the average force exerted by the boy on the ball.
I hope this helps! Let me know if you have any further questions.