Question

A batter hits a pitched ball when the center of the ball is 1.28 m above the ground.The ball leaves the bat at an angle of 45° with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m. (a) Does the ball clear a 8.22-m-high fence that is 97.0 m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

Answer 1

PROJECTILE

1)

along horizontal
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initial velocity v0x = v*cos(theta)

acceleration ax = 0

initial position = xo = 0

final position = x = 107

displacement = x - xo

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = v*costheta*T

T = (x - x0)/(v*cos(theta))......(1)

along vertical
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v0y = v*sin(theta)

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 1.28 m

final position y = 0

from equation of motion

y-y0 = vy*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = (v*sin(theta)*(x-x0))/(v*cos(theta)) - (0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2)

y-y0 = (x-x0)*tan(theta) - ((0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2))

0-1.28 = (107-0)*tan45 - (0.5*9.8*(107-0)^2/(v^2*(cos45)^2))

launch speed v = 32.2 m/s

part(a)

along horizontal   for x = 97 m

x - x0 = v*costheta*t

97 - 0 = 32.2*cos45*t

time taken t = 4.26 s

along vertical at time t = 4.26 s

y - y0 = v*sintheta*t + (1/2)*ay*t^2

y - 1.28 = 32.2*sin45*4.26 - (1/2)*9.8*4.26^2

y = 9.35 m   which greater than 8.22 m

the ball will clear the fence

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part ( b)

distance between fence top and ball center = 9.35 - 8.22 = 1.13 m