Answer to Question #242333 in Physics for kathleen

Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second snowball at a low angle and time it to arrive at the same time as the first.

Assume both snowballs are thrown with

the same initial speed 28.2 m/s. The first

snowball is thrown at an angle of 63◦ above

the horizontal. How many seconds after the first snowball should you throw the second so that they arrive on target at the same time?

Answer in units of s

Answer 1

The time of the flight of the first snowball:

$$t_1=\frac {v\sin\theta_1}{g}.$$

The time for the second ball to pass the same distance:

$$t_2=\frac {v\sin\theta_2}{g}=t_1+\Delta t.\\\space\\ \frac {v\sin\theta_1}{g}=\frac {v\sin\theta_2}{g}+\Delta t,\\\space\\ \Delta t=\frac vg(\sin\theta_1-\sin\theta_2)=2.88(0.891-\sin\theta_2)\text{ s},$$

where $\theta_2$ - the angle of launch for the second snowball.