A baseball is thrown downward 8


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.

A baseball is thrown downward

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Solution:

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We will use the following formula to solve this question to figure out the speed at which it hits the ground.

v_2^2=v_1^2+2a_c(s_2-s_1)

(Where v_2 is final velocity, v_1 is initial velocity, a_c is constant acceleration, s_2 is initial displacement, and s_1 is final displacement)

 
From the question, we know the following values:

v_1=18 ft/s

a_c=32.2 \,\text{ft/s}^2

s_2=50 ft

s_1=0 ft

 

Let us substitute these values into our equation:

v_2^2=v_1^2+2a_c(s_2-s_1)

v_2^2=18^2+2(32.2)(50-0)

v_2^2=3544

(Take the square root of both sides)

v_2=59.5 ft/s

 

To figure out the time of travel, we will use the following equation.

v_2=v_1+a_ct

(Where v_2 is final velocity, v_1 is initial velocity, a is constant acceleration and t is time)

 

Again, substituting the values we know gives us:

v_2=v_1+a_ct

59.5=18+32.2(t)

(Solve for t)

t=1.29 s

 

Final Answers:

v_2=59.5 ft/s

t=1.29 s

 

This question can be found in Engineering Mechanics: Dynamics, 13th edition, chapter 12, question 12-1.

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