Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal. How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?

Answer 1

Answer:

3.60342 seconds

Explanation:

v = Initial velocity of snowball = 25 m/s

g = Acceleration due to gravity = 9.81 m/s²

$\theta_1$ = First angle = 75°

The second angle will be

$\theta_2=90-\theta_1=90-75=15^{\circ}$

Horizontal speed for first throw

$v_{x1}=vcos\theta_1\\\Rightarrow v_{x1}=25\times cos75\\\Rightarrow v_{x1}=6.47047\ m/s$

Horizontal speed for second throw

$v_{x2}=vcos\theta_1\\\Rightarrow v_{x2}=25\times cos15\\\Rightarrow v_{x2}=24.14814\ m/s$

Horizontal range is given by

$R=\frac{v^2sin(2\theta)}{g}\\\Rightarrow R=\frac{25^2\times sin(2\times 75)}{9.81}\\\Rightarrow R=31.85\ m$

Time period for first throw

$t_1=\frac{R}{v_{x1}}\\\Rightarrow t_1=\frac{31.85}{6.47047}\\\Rightarrow t_1=4.92236\ s$

Time period for second throw

$t_2=\frac{R}{v_{x1}}\\\Rightarrow t_2=\frac{31.85}{24.14814}\\\Rightarrow t_2=1.31894\ s$

The time difference is

$t=4.92236-1.31894=3.60342\ s$

The second ball should be thrown 3.60342 seconds later