   Chapter 3, Problem 57AP

Chapter
Section
Textbook Problem

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that 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 25.0 m/s. The first is thrown at an angle of 70.0° 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 for both to arrive at the same time?

To determine
The angle with which the second snowball be thrown to arrive at the same point as the first.

Explanation

For the vertical motion of the first snowball,

Δy=v0yt+12ayt2=(v0sinθ1)t1+12ayt12

Substitute 0 for Δy , 25.0m/s for v0 , 70.0° for θ and 9.80m/s2 for ay to find the time t1 .

0=[(25.0m/s)sin70.0°]t1+12(9.80m/s2)t12t1=2(25.0m/s)sin70.0°9.80m/s2=4.79s

The horizontal displacement for the first snowball is,

Δx=v0xt1=(v0cosθ)t1

Substitute 25.0m/s for v0 , 70.0° for θ and 4.79s for t1 .

Δx=[(25.0m/s)cos70.0°]4.79s=41.0m

For the vertical motion of the second snowball,

Δy=v0yt+12ayt2=(v0sinθ2)t2+12ayt22

Substitute 0 for Δy , 25

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