Question

2. A ball is thrown at an angle of 45.0° from the horizontal with an initial velocity of 5.0 𝑚/𝑠.

a. At what time will the ball reach its maximum height?

b. What will be the maximum height of the ball’s trajectory?

c. How far will the ball travel horizontally when it reaches the maximum height of its flight?

Answer 1

a. At what time will the ball reach its maximum height?

$$t_1=\frac{v_0\sin\theta}{g}=\frac{5.0\sin45^\circ}{9.8}=0.36\:\rm s$$

b. What will be the maximum height of the ball’s trajectory?

$$h_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{5.0^2\sin^245^\circ}{2*9.8}=0.63\:\rm m$$

c. How far will the ball travel horizontally when it reaches the maximum height of its flight?

$$l=v_0\cos\theta*t_1=5.0\cos45^\circ*0.36=1.27\:\rm m$$