Question

A projectile has an initial velocity of 110 m/s and a launch angle of 40o from the horizontal. The surrounding terrain is level, and air friction is to be disregarded. What is the maximum elevation achieved by the projectile?

Answer 1

255.09 m

Step-by-step explanation:

Using

H = U²sin²θ/2g.................. Equation 1

Where H = maximum elevation, U = initial velocity, θ = Angle of projectile, g = acceleration due to gravity.

Given: U = 110 m/s, θ = 40°, g = 9.8 m/s².

Substitute into equation 1,

H = 110²sin²40/(2×9.8)

H = 12100(sin²40)/(19.6)

H = 12100(0.4132)/19.6

H = 255.09 m.

Hence the maximum elevation achieved by the projectile = 255.09 m