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 one 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 16.1 m/s. The first one is thrown at an angle of 72◦ with respect to the horizontal. At what angle should the second snowball be thrown to arrive at the same point as the first? Answer in units of ◦

Answer 1

`\theta_2 \approx 40.5^(\circ)`

Step-by-step explanation:

Given:

velocity of the ball first and ball second,
`v_1=v_2=v\,m.s^(-1)`

angle of projection of the first ball,
`\theta_1= 72^(\circ)`

∵The balls should land at the same point,

∴their range of projectile,

`R_1=R_2=R\,m`

As we know for the range of projectile:

`R=(v^2)/(g).sin\,2\theta`

∵ we have equal range in both the cases

`\therefore (v_1^2)/(g).sin\,2\theta_1=(v_2^2)/(g).sin\,2\theta_2`

`\Rightarrow (v^2)/(g).sin\,(2* 72) =(v^2)/(g).sin\,2\theta_2`

`\Rightarrow sin 144^(\circ)=sin\,2\theta_2`

`2\theta_2=sin^(-1)[sin 144^(\circ)]`

`\theta_2=(1)/(2)* sin^(-1)[sin 144^(\circ)]`

`\theta_2 \approx 40.5^(\circ)`