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

3.60342 seconds

Step-by-step 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* cos75\\\Rightarrow v_(x1)=6.47047\ m/s`

Horizontal speed for second throw

`v_(x2)=vcos\theta_1\\\Rightarrow v_(x2)=25* cos15\\\Rightarrow v_(x2)=24.14814\ m/s`

Horizontal range is given by

`R=(v^2sin(2\theta))/(g)\\\Rightarrow R=(25^2* sin(2* 75))/(9.81)\\\Rightarrow R=31.85\ m`

Time period for first throw

`t_1=(R)/(v_(x1))\\\Rightarrow t_1=(31.85)/(6.47047)\\\Rightarrow t_1=4.92236\ s`

Time period for second throw

`t_2=(R)/(v_(x1))\\\Rightarrow t_2=(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