Question

William throws the ball to Caleb who is standing 40 meters away. 1/4 of the way the ball is going Caleb it’s height is at 8 meters. Calculate the max height.

Answer 1

Solution. Consider the general equations of motion at an angle where the movement in the horizontal and vertical direction is independent. In the horizontal direction, the body moves at a constant speed.

$$v_x=vcos\alpha$$

where v is initial velocity of the ball.

In the vertical direction, the body moves with an initial upward curvature and the equation of skrasty and height can be written as

$$v_y=vsin\alpha - gt$$

$$h=vtsin\alpha -\frac {gt^2} {2}$$

where g=9.8 m/s^2 is constant gravity; t is time. According to the condition of the problem

$$vt_1cos\alpha=10$$

$$8=vt_1sin\alpha -\frac {gt_1^2} {2}$$

Therefore

$$t_1=\frac {10} {vcos\alpha}$$

$$8=10tan\alpha -\frac {100g} {2v^2cos^2\alpha}$$

The ball will be at the maximum height at exactly half of its trajectory. Hence time is equal to

$$t_2=\frac {20} {vcos \alpha}.$$

On the other hand, the maximum lift height corresponds to zero vertical speed.

$$0=vsin\alpha - gt_2$$

$$t_2=\frac {vsin\alpha} {g}$$

Therefore

$$v^2 sin \alpha cos \alpha =20g$$

As result

$$8=10tan\alpha -\frac {5v^2 sin \alpha cos \alpha} {2v^2cos^2\alpha}$$

$$8=10tan\alpha-\frac {5} {2} tan\alpha \Longrightarrow tan\alpha=\frac {16} {15}$$

The maximum lift height will be found using the formula

$$h_{max}=\frac {v^2sin^2 \alpha} {2g}$$

Hence

$$h_{max}=\frac {v^2sin^2 \alpha} {\frac {v^2 sin \alpha cos \alpha} {10}}=\frac {10v^2sin^2 \alpha} {v^2 sin \alpha cos \alpha}=10tan\alpha$$

$$h_{max}=10 \times \frac {16} {15}=\frac {32} {3} meters$$

Answer.

$$h_{max}=\frac {32} {3} meters$$