Question

A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.

Answer 1

• Initial velocity=16m/s=u
• Angle=33°
• Acceleration due to gravity=9.8m/s^2=g

We have to find Range

$\\ \bull\tt\dashrightarrow R=\dfrac{u^2sin(2\Theta)}{g}$

$\\ \bull\tt\dashrightarrow R=\dfrac{16^2sin(2\times 33)}{9.8}$

$\\ \bull\tt\dashrightarrow R=\dfrac{256sin66}{9.8}$

$\\ \bull\tt\dashrightarrow R=\dfrac{256(0.91)}{9.8}$

$\\ \bull\tt\dashrightarrow R=\dfrac{232.96}{9.8}$

$\\ \bull\tt\dashrightarrow R=23.77m$

Answer 2

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m