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 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 that 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 in order for both to arrive at the same time?____________________

Explanation / Answer

X1 = xo1 + vox1t1 + (1/2)ax1t12
X1 = 0 + (20.0m/s)cos(75o)t1 + 0
Y1 = yo1 + voy1t1 + (1/2)ay1t12
0m = 0m + (20.0m/s)sin(75o)t1 + (1/2)(-9.81m/s2)t12
t1 = (20.0m/s)sin(75o)/(4.905m/s2) = 3.94s
X1 = (20.0m/s)cos(75o)(3.94s) = 20.4m
X2 = xo2 + vox2t2 + (1/2)ax2t22
X1 = X2
20.4m = 0m + (20.0m/s)cos()t2 + 0
t2 = (20.4m)/[(20.0m/s)cos()] = 1.02cos-1() s
Y2 = yo2 + voy2t2 + (1/2)ay2t22
0m = 0m + (20.0m/s)sin()t2 - (4.905m/s2)t22
t2 =(20.0m/s)sin()/(4.905m/s2) = 4.08sin() s

t2 = 1.02cos-1() s = 4.08sin() s
(1.02s)/(4.08s) = sin()cos()
Use trig. id.
sin(2) = 2sin()cos()
sin()cos() = (1/2)sin(2)
(1.02s)/(4.08s) = (1/2)sin(2)
2 = sin-1[2(1.02s)/4.08s] = sin-1[0.50] = 30o
= (1/2)30o = 15o
t2 = 4.08sin() s = 4.08sin(15o) s = 1.06s
tthrow = t1 - t2 = (3.94 - 1.06)s = 2.88s

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