Question

A cannon ball is fired with an initial velocity of 75 m/s at an angle of 58 degree above the ground. What maximum height will it reach and how far it fly horizontally

Answer 1

Answer:

h = 206.4 m

range = 515.9 m

Explanation:

from the question we are given the following:

initial velocity (u) = 75 m/s

angle  above surface = 58 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

find the maximum height (h) and the horizontal distance

maximum height (h) = $\frac{u^{2}sinθ^{2} }{2g}$

h =  $\frac{75^{2}sin58^{2} }{2 x 9.8}$

h = 206.4 m              

the horizontal distance here is the range

range = $\frac{u^{2}sin2θ }{g}$

range = $\frac{75^{2}sin(2 x 58) }{9.8}$

range = 515.9 m