One strategy in a snowball fight is to throw a snowball at a high angle over lev

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 36 m/s. The first one is thrown at an angle of 54? 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 degrees.

Explanation / Answer

For first ball: 0 = 20sin(65)-9.81t =>t = 18.12615574/9.81 = 1.847722298sec x = 20cos(65)*2*1.847722298 = 31.23524743m Let second ball be thrown at an angle theta. Then 0 = 20sin(theta)-9.81t' =>t' = 20sin(theta)/9.81 x = 20cos(theta)*2*20sin(theta)/9.81 = 31.23524743 =>(400/9.81)*sin(2theta) = 31.23524743 =>sin(2theta) = 0.766044443 =>2Theta = 50 =>Theta = 25 Degrees So the second ball is thrown at an angle 25 degrees to the horizontal. t' = 20sin(theta)/9.81 = 0.861607058sec t-t' = 1.847722298-0.861607058 = 0.98611524sec So it should be thrown after 0.98611524sec.

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