Question

An object is fired from the ground at 275 m/s at an angle of 55 degrees above the horizontal. How far away did the object first hit the ground? what is the maximum height that the object reaches?

Answer 1

The diagram representing this scenario is shown below

The object is undergoing projectile motion. To find the distance between the object's initial position and the point where it first hit the ground, we would apply the formula for calculating Range which is expressed as

Range = U^2Sin2θ/g

where

U is the initial velocity of the object = 275

θ = angle made with the horizontal = 55

g = acceleration due to gravity = 9.8

By substituting these values into the formula,

Range = 275^2 x Sin(2 x 55)/9.8 = 75625 x Sin110/9.8 = 7251.5

distance between the object's initial position = 7251.5 m

To find the maximum height H, we would apply the formula which is expressed as

H = V^2Sin^2θ/2g

V = 275

θ = 55

g = 9.81

H = 275^2 x 2 x (Sin 55)^2/2 x 9.8 = 5178.1

the maximum height that the object reaches = 5178.1 m