Question:

A ball is thrown horizontally from the height of 40.0 m. The horizontal range of the projectile is known as 80 .0 m. What is the angle that the velocity vector makes with the horizontal right before the ball hits the ground?

Answer:

The time to ground is obtained from the \(y\)-direction of the motion. Namely,

\[ y = -\frac{1}{2}gt^2 \]

Solve for \(t\) and plug in the numbers.

\[  t = \sqrt{\frac{-2y}{g}}  \]

\[  = \sqrt{\frac{-2\cdot (-40)}{9.8}}   \]

\[  = 2.857 \mathrm{s}  \]

Then, we can find the velocity of \(x\)-direction.

\[ v_x = \frac{x}{t} = \frac{80}{2.857} = 28.0 \mathrm{m/s}\]

The velocity of \(y\)-direction right before hitting ground is

\[ v_y = -gt = -9.8 \times 2.857 = -28.0 \mathrm{m/s} \]

Then, the angle is calculated as follows:

\[ \theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) = \tan^{-1} \left(\frac{-28.0}{28.0}\right) = 315^o\]