Question

A ball is thrown at an angle 40.00 above the horizontal with an initial velocity of 22.0 m/s. What is the range of the ball?

Answer 1

48.64 m

Step-by-step explanation:

From the question,

Range(R) = (U²Sin2Ф)/g.................. Equation 1

Where U = initial velocity, Ф = Angle to the horizontal, g = acceleration due to gravity.

Given: U = 22 m/s, Ф = 40°, g = 9.8 m/s².

Substitute these values into equation 1

R = 22²Sin(40×2)/9.8

R = 484×0.9848/9.8

R = 48.64 m

Hence the range of the ball is 48.64 m