Question

A ball was kicked at an angle of 15 degree above the horizontal. The speed is 30 m/s at horizontal dimension (initial). Neglecting air resistance, the projectile will travel at what horizontal distance before striking the ground? Hint: Find first the time before its strikes the ground. Bonus: Get the time as it reaches its peak, the maximum height.

Answer 1

Given:

$\theta=15^\circ$

$v_0=30\:\rm m/s$

$g=9.8\:\rm m/s^2$

The range of projectile

$$R=\frac{v_0^2\sin2\theta}{g}=\rm\frac{(30\:m/s)^2\sin30^\circ}{9.8\: m/s^2}=46\: m$$

The time of movement to the highest point of trajectory

$$t=\frac{v_0\sin\theta}{g}=\rm\frac{30\: m/s*\sin15^\circ}{9.8\: m/s^2}=0.79\: s$$