Question

A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maximum height at point p then falls to point a, which is 65 m below the cliff. What is the horizontal distance that it travels (labeled x in the diagram)

Answer 1

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity =$v_{0y}(t)$ = 25 m/s* sin(30°)= 12.5 m/s

height = h

$h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s$

t- time is found solving quadratic equation.

horizontal velocity = $v_{0x}=25m/s*cos(30^{o})=21.65 m/s$

Horizontal velocity is constant, so distance $x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m$