Question

Example 1: A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel.

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Answer #1

Given,

Initial height of the baseball (or the tower);

Yo=50 ft

Initial velocity (downwards);

Vo=-18 ft/s

Acceleration is acting downwards due to gravity;

a_y = -g = -32.2;;ft/s^2

Final position of the ball when it hits the ground level;

y=0

Vertical displacement;

Delta s_y = y-y_o

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By third equation of motion;

v^2 = v_o^2 + 2a_yDelta s_y

Rightarrow v^2 = v_o^2 + 2a_y(y-y_o)

수 1,2 = (-18)2+2(-32.2)(0-50)

1,2 = (-18)-+2(-32.2)(-50)

u= 59.53 ft/s (.)

...(Answer)

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Second equation of motion is;

y = y_o +v_ot + rac{1}{2}a_yt^2

Rightarrow 0 = 50 +(-18)t + rac{1}{2}(-32.2)t^2

Rightarrow 16.1t^2+18t - 50 = 0

18+182- 4(16.1) (-50) 2(16.1)

18+182 +4(16.1) (50) 2(16.1)

Rightarrow t =1.2898;;s

...(Answer)

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