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?

Please show it step by step along with the equation to use

I still don't get it. Can you plug the numbers in and tell me which equation to use

force=.15*11/1.5=?

Is this the equation: F=ma/?

The answer I'm suppose to get is 7.55N

How do I get that?

To find the average force exerted by the boy on the ball, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. The equation is:

F = m * a

To find the force, we need to determine the acceleration of the ball. Since the ball is thrown straight up into the air and it eventually comes to a stop at the highest point, we know that the acceleration of the ball is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2 (assuming no air resistance).

The initial velocity of the ball is given as 11 m/s. Since the ball is thrown straight up, its final velocity at its highest point is 0 m/s. We can use the following equation to relate the initial velocity, final velocity, acceleration, and displacement:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s)
u = initial velocity (11 m/s)
a = acceleration (-9.8 m/s^2)
s = displacement (1.5 m)

Rearranging this equation to solve for acceleration (a):

a = (v^2 - u^2) / (2s)

a = (0 - 11^2) / (2 * 1.5)
a = -121 / 3

Now that we have the acceleration, we can substitute it into Newton's second law equation:

F = m * a
F = 0.15 kg * (-121 / 3)

Calculating the force:

F ≈ -7.65 N

Therefore, the average force exerted by the boy on the ball is approximately -7.65 Newtons. The negative sign indicates that the force is directed downward, opposing the upward motion of the ball.

impulse=changemomentum

force*1.5=.15*11
solve for force.