Question

a cannon ball is fired at 40 degrees with an initial velocity of 180 m/s. what is the maximum height reached by the cannonball?

Answer 1

A ball is fired with speed v = 180 m/s at an angle 40 degree

So here we can say the components of velocity is given as

$v_{x} = 180 cos40 = 137.9 m/s$

$v_y = 180 sin40 = 115.7 m/s$

now when ball will reach the maximum height its velocity in vertical direction will become zero

so here we can say

$v_{fy} = 0$

now using kinematics we will have

$v_{fy}^2 - v_y^2 = 2 a d$

$0 - 115.7^2 = 2(-9.8)(H)$

so maximum height is

$H = 683 m$

so it will reach at maximum height of 683 m