physics help?

Question

A baseball is hit at an initial velocity of 110 ft/s at an angle of 45 degrees above the horizontal. Will the ball clear a 10.0 ft fence 300 ft from home plate for a home run? If so, by how much will it clear the fence? If not, how far in front of the fence will it land or where will it hit the wall?

Answer 1

Assume the ball was hit at ground level or a height of 3 ft?
I will use 3 ft

y(t)=3+110*sin(45)*t-16*t^2
for y(t)=10.0, find t
-7+110*SIN(45)*t-16*t^2

The root of interest is 4.77 Seconds

What is x(4.77)?
cos(45)*110*4.77
371 ft, it clears the wall by a good bit

x(t)=300=110*sin(45)*t
solve for t
t=3.857

y(3.857) is
65 ft. The ball clears by 54 ft.

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Answer 2

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The equation of path of projectile is:

y = x tan theta -gx^2/2u^2cos^2theta

Given , x= 300 ft

theta =45 degree

y = 300 -32*300*300/110*110 = 61.98347 ft

The height of the ball is 61.98347 feet at 300 feet from point at which it is hit.

The ball will clear the fence.

The ball will clear the fence by 51.98347 feet.
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The range of base ball =R=u^2 sin 2theta / g

R =110*110*sin90 /32 =378.125 ft

the ball will clear the fence land 78.125 feet away from the fence
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Answer 3

ur mom