In a baseball game, the ball hits the top of a wall that is 26 m high and located at 150m away from the thrower, if the
ball is hit at an angle 30degree(0)

to the horizontal, assume that the ball is hit at a height of 3 m above the ground. Find :
a. The initial speed of the ball
b. The time it takes the ball to reach the wall
c. The velocity components and the speed of the ball when it reached the wall

When the ball is hit at speed v, the horizontal component is v·cos(30°), and the vertical component is v·sin(30°). Then the position as a function of time is ...

(x, y) = (v·cos(30°)·t, -4.9t² +v·sin(30°)·t +3)

We want to find v and t such that (x, y) = (150, 26).

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Solution

Using the value of x first, we find ...

vt·cos(30°) = 150

vt = 150/cos(30°) = 100√3

Substituting this into the relation for y, we have ...

-4.9t² +vt·sin(30°) +3 = 26

-4.9t² +50√3 -23 = 0

t² -12.980110 = 0 . . . . . divide by -4.9

t ≈ 3.603 . . . . . . . . . . . . add the constant, take the square root

a. Then the initial speed of the ball is ...

vt = 100√3

v = 100√3/t ≈ 48.08 . . . . m/s

The initial speed of the ball is about 48.08 m/s.

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b. The time it takes to reach the wall is 3.603 seconds.

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c. The horizontal velocity component is ...

v·cos(30°) ≈ 41.63 m/s . . . horizontal velocity

The vertical component of velocity at the wall will be found from the derivative of the vertical position expression;

y' = -2(4.9)t +v·sin(30°) = -9.8(3.603) +48.075/2 = -11.27 . . . m/s

The horizontal and vertical velocity components at the wall are ...

(x', y') ≈ (41.63, -11.27) m/s