Question

A projectile is launched at an angle 60° to the horizontal
with some initial speed vi-45 m/s, and air resistance is
negligible. The cliff is 265 m high.
A) Find the final velocity as a vector,
B) Find the maximum height.
C) Find the horizontal range of the motion.

Answer 1

Explanation:

We have,

A projectile is launched at an angle 60° to the horizontal  with some initial speed 45 m/s.

The cliff is 265 m high.

(A) The final velocity of the projectile is given by :

$v_f^2-v_i^2=2gs\\\\v_f^2=2gs+v_i^2\\\\v_f^2=2\times 9.8\times 265+45^2\\\\v_f=84.96\ m/s$

(B) The maximum height of the projectile is given by :

$H=\dfrac{v_i^2\sin^2\theta}{2g}\\\\H=\dfrac{45^2\times \sin^2(60)}{2\times 9.8}\\\\H=77.48\ m$

(C) The  horizontal range of the motion is given by :

$R=\dfrac{v_i^2\sin 2\theta}{g}\\\\R=\dfrac{(45)^2\times \sin 2(60)}{9.8}\\\\R=178.94\ m$