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chubhunter [2.5K]
3 years ago
5

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

irst 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 20.0 m/s. The first one is thrown at an angle of 75.0° 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?
Physics
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

Part a)

\theta_2 = 15 degree

Part b)

\Delta t = 2.88 s

Explanation:

Part a)

In order to have same range for same initial speed we can say

R_1 = R_2

\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}

so after comparing above we will have

\theta_1 = 90 - \theta

so we have

75 = 90 - \theta_2

\theta_2 = 15 degree

Part b)

Time of flight for the first ball is given as

T_1 = \frac{2vsin\theta}{g}

T_1 = \frac{2(20)sin75}{9.81}

T_1 = 3.94 s

Now for other angle of projection time is given as

T_2 = \frac{2(20)sin15}{9.81}

T_2 = 1.05 s

So here the time lag between two is given as

\Delta t = T_1 - T_2

\Delta t = 3.94 - 1.05

\Delta t = 2.88 s

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