Question

A ball is thrown with an initial velocity of 8m/s in a direction that makes an angle of 35° with the horizontal. Find the horizontal range of the ball?

Answer 1

Answer:

The horizontal range of the ball is 6.13 m

Step-by-step explanation:

The initial velocity of the ball, u = 8 m/s

Angle made by the ball with the horizontal, θ = 35°

Acceleration due to gravity, g = 9.81 m/s²

The horizontal range is calculated as shown below

`R=(u^2\sin 2\theta)/(g)`

Substitute u = 8 m/s, θ = 35°, and g = 9.81 m/s² into the range formula above

`\begin{gathered} R=(8^2\sin (2*35))/(9.81) \\ R=(64\sin (70))/(9.81) \\ R=(64(0.9397))/(9.81) \\ R=6.13\text{ m} \end{gathered}`

Therefore, the horizontal range of the ball is 6.13 m