A football is kicked at 37degrees to the horizontal and it goes a distance of 98 yd. Ignore air resistance. a) what is the ball’s initial velocity? b) what is the vertical component of the ball’s velocity? c) what is the horizontal component of the ball’s velocity d) how long is the ball in the air? (Choose positive vertical direction upwards)

Question

If the same football was punted with the same initial velocity (from question a) on the surface of the moon where a=1.62m/s^2 how long does it take the ball to return to the ground? What is the max height? What is the horizontal range?

98yd=89.61m

a) delta x=v_0^2/g
sqrt((89.61)(10))=29.9m/s
b) voy=v_0sin theda
=(29.9)sin37
=17.99
c) vox=v_0cos theda
=(29.9)cos37
=23.87m/s
d) t=d/vx
=89.61/23.87
=3.75s

For where the moon has a=1.62m/s^2, wouldn’t time also equal 3.75s
for the max height:
d =voy(t) +1/2at^2
=17.99*1.875=1/2 (-1.62)(1.875)^2
=33.73+ -2.84
=30.89m
(would it be -1.62 or would it be positive if we choose vertical direction to be up?)

the range would be:
delta x=vx(t)
=(23.87)(3.75)
=89.51m

Are these correct?

Answer 1

A football is kicked at 37 degrees to the horizontal and it goes a distance of 98 yd. Ignore air resistance. a) what is the ball’s initial velocity? b) what is the vertical component of the ball’s initial velocity? c) what is the horizontal component of the ball’s initial velocity d) how long is the ball in the air? (Choose positive vertical direction upwards)

a--The horizontal distance traveled is 98(3) = 294 feet.
d = V^2(sin2µ)/g where d is the horizontal distance traveled in feet, V is the initial velocity in ft./sec., µ is the velocity angle with the horizontal and g = the acceleration due to gravity, 32.2 ft./sec.^2.
Therefore, using d = V^2sin(2µ)/g, 294 = V^2(sin74)/32.2 = from which Vo = 99.24 ft/sec.

b--The vertical component of the initial velocity is 99.24(sin37) = 59.72 ft/sec. ft/sec.

c--The horizontal component of the initial velocity is 99.24(cos37) = 79.25 ft/sec.

d--From Vf = Vo - gt where Vo = the initial vertical velocity component, Vf = the final vertical velocity component (= 0) and t = the time from time = 0 to the maximum height at t(up), 0 = 99.23 - 32.2t(up), t(up) = 99.23/32.2 = 3.08 sec.
Since it takes the same amount of time to fall back to the ground, the total time in the air is 6.16 sec.

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Given the above, you should be able to determine the moon punt answers.