Question

A baseball player hits a home run over the left-field fence, which is 104 m from...

A baseball player hits a home run over the left-field fence, which is 104 m from home plate. The ball is hit at a point 1.07 m directly above home plate, with an initial velocity directed 28.0° above the horizontal. By what distance does the baseball clear the 3.00 m high fence, if it passes over it 3.15 s after being hit?

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Answer #1

Given is:-

Initial height y0 = 1.07m

Time t=3.15s

horizontal distance x = 104m

Now,

Let the initial velocity be = u m/s

and its components are

uucos(28)m/s, uy- usin(28)m/s

Now,

The final displacement in y direction can be found out by using second equation of motion, thus

y = y_o + u_y t + rac{1}{2}at^2

by plugging all the values we get

.07(usin(28)(3.15)+9.81(3.15)2 eq-1

Now,

the horizontal distance will be

d = u_x imes t

by plugging all the values we get

104 ucos(28) (3.15)

which gives us

u-37.4m/s

by plugging this value in eq-1 we get

y = 1.07 + (37Asin(29) (3.15) +ー(-9.81) (3.15)2

y = 7.71m

The difference between baseball and the fence is

y' =y-y_o

7.71-3

y 4.71m

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