Question

A football is kicked off the flat ground at 23.0 m/sat an angle of 30∘ relative to the ground.

Find the angle of its velocity with respect to the ground after it has been in the air for one-fourth of this time.

Repeat for one-half of the total time.

Repeat for three-fourths of the total time.

For each of these times, determine its speed.

Answer #1

t = 2 v sin(theta) / g = (2 x 23 x sin30) / 9.8

t = 2.35 sec

v0x = 23 cos30 = 19.92 m/s

v0y = 23 sin30 = 11.5 m/s

for t1 = t/4 =0.5875 sec

vx = v0x = 19.92 m/s

vy = v0y + ay t = (11.5) + (-9.8 x 0.5875)

vy = 5.74 m/s

Speed = sqrt(vx^2 + vy^2) = 20.7 m/s ....Ans

angle = tan^-1(vy/vx) = 16 deg ....Ans

for t1 = t/2 = 1.175 sec

vx = v0x = 19.92 m/s

vy = v0y + ay t = 0

Speed = sqrt(vx^2 + vy^2) = 19.92 m/s ....Ans

angle = tan^-1(vy/vx) = 0 deg ....Ans

for t3 = 3 t / 4

Speed = 20.7 m/s

angle = - 16 deg ( minus indicates that it is below the horizontal)

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