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

h = 206.4 m

range = 515.9 m

Step-by-step 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) =
`(u^(2)sinθ^(2)  )/(2g)`

h =
`(75^(2)sin58^(2)  )/(2 x 9.8)`

h = 206.4 m

the horizontal distance here is the range

range =
`(u^(2)sin2θ )/(g)`

range =
`(75^(2)sin(2 x 58) )/(9.8)`

range = 515.9 m