A 1.0kg football is given an initial velocity at ground level of 20.0m/s (37 degrees above horizontal). it gets blocked just after reaching a height of 3.0m
Using conservation of energy:
Initial Energy = Final Energy
Initial Energy = KE + PE
Final Energy = KE + PE
Since the football is at ground level initially, the initial potential energy is zero:
Initial Energy = KE
Initial Energy = 0.5mv^2
Initial Energy = 0.5(1.0kg)(20.0m/s)^2
Initial Energy = 200.0 J
At its maximum height of 3.0m, the football's potential energy is:
PE = mgh
PE = (1.0kg)(9.81m/s^2)(3.0m)
PE = 29.43 J
Using trigonometry, we can find the vertical component of the football's velocity when it reaches its maximum height:
v_y = v*sin(37)
v_y = 20.0m/s*sin(37)
v_y = 12.1 m/s
At the maximum height, the football's kinetic energy is zero, so the final energy is just the potential energy:
Final Energy = PE
Final Energy = 29.43 J
Since no external work is done on the football, the initial energy must equal the final energy:
Initial Energy = Final Energy
0.5mv^2 = mgh
0.5v^2 = gh
v^2 = 2gh
v = sqrt(2gh)
v = sqrt(2*9.81m/s^2*3.0m)
v = 7.67 m/s
This is the speed at the maximum height, but we want to find the speed just as the football is blocked. We can use conservation of momentum to find this final velocity.
Since the football is being blocked, its final momentum is zero. Therefore, the initial momentum must also be zero:
Initial Momentum = Final Momentum
p_i = p_f
mv_i = 0
v_i = 0
Now we can use conservation of energy to find the velocity just as the football is blocked:
Initial Energy = Final Energy
KE_i + PE_i = KE_f + PE_f
0.5mv_i^2 + mgh_i = 0.5mv_f^2 + mgh_f
0.5(1.0kg)(20.0m/s)^2 + (1.0kg)(9.81m/s^2)(0m) = 0.5(1.0kg)(v_f)^2 + (1.0kg)(9.81m/s^2)(3m)
200 J = 0.5v_f^2 + 29.43 J
170.57 J = 0.5v_f^2
v_f^2 = 341.14
v_f = 18.47 m/s
Therefore, the final velocity of the football just as it is blocked is 18.47 m/s.