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 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 70.0 degree with respect to the horizontal, (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?

Explanation / Answer

here,

initial speed , u = 25 m/s

angle for first snowball , theta1 = 70 degree

a)

the horizontal range is same for the angle theta and (90 - theta)

so, the angle of launch for seccond snowball , theta2 = 90 - theta1

theta2 = 20 degree

b)

time of flight for 1 , t1 = 2 * u * sin(theta1)/g

t1 = 2 * 25 * sin(70) /9.8 = 4.79 s

time of flight for 2 , t2 = 2 * u * sin(theta2)/g

t2 = 2 * 25 * sin(20) /9.8 = 1.75 s

the seccond snowball may be thrown after (4.79 - 1.75 = 3.04 s)

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