Answer to Question #158455 in Physics for ivanna

Question

A racquetball thrown from the ground at an angle of 28° and with a speed of 35 m/s lands exactly 1.6 s later on the top of a nearby building. Calculate the horizontal distance it traveled and velocity just before hitting the building.

Answer 1

The horizontal distance it traveled:

$$R=v_xt=v\text{ cos}28°t=49\text{ m}.$$

Find the velocity just before hitting the building. First, calculate the time it takes to reach the highest point of trajectory:

$$\tau=\frac{v_y}{g}=\frac{v\text{ sin}28°}{g}=1.6767\text{ s}.$$

It means that after 1.6 s, the object had certain vertical component of velocity, which equals

$$v_{y1}=v_y-gt=v\text{ sin}28°-gt=0.75\text{ m/s}.$$

So, the magnitude of velocity right before hitting the building is

$$v(t)=\sqrt{v_x^2+v_{y1}^2}{}=30.9\text{ m/s}.$$