 #1
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Homework Statement:

Ball (mass m) launched horizontally at height H from ground.
Horizontal displacement at first bounce is 4/3 H and maximum height after the first bounce is 9/16 H.
Duration of contact between ball and ground is t
Homework Equations:
 
Here's what I have done:
Let t1 be the time taken for the ball to reach the ground after the first bounce, and ux be the initial speed in the horizontal direction of ball.
Also, let v be the speed after the bounce.
For horizontal,
$$\dfrac {4}{3}H=U_{x}t_{1} $$
For vertical,
$$H=\dfrac {1}{2}g\left( t_{1}\right) ^{2} $$
Solving these 2, I find:
$$U_{x}=\dfrac {2}{3}gt_{1} $$
Im supposed to find the vertical average force. In this case will it be:
$$F_{avg}=\dfrac {m\left( v_{y}u_{y}\right) }{t} $$
?
Not sure what the question is requiring me to do
Let t1 be the time taken for the ball to reach the ground after the first bounce, and ux be the initial speed in the horizontal direction of ball.
Also, let v be the speed after the bounce.
For horizontal,
$$\dfrac {4}{3}H=U_{x}t_{1} $$
For vertical,
$$H=\dfrac {1}{2}g\left( t_{1}\right) ^{2} $$
Solving these 2, I find:
$$U_{x}=\dfrac {2}{3}gt_{1} $$
Im supposed to find the vertical average force. In this case will it be:
$$F_{avg}=\dfrac {m\left( v_{y}u_{y}\right) }{t} $$
?
Not sure what the question is requiring me to do