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Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s2, how high above the ground is the ball 2 s after it is thrown?

User Sheina
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2 Answers

5 votes
hello

the asnwer is 12 feet

have a nice day
User Rajalakshmi
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4 votes

Explanation:

It is given that, two boys are throwing a baseball back and forth.

Height above ground, h₀ = 4 ft

Initial velocity of child, u = 36 ft/s

Acceleration due to gravity, a = -16 ft/s²

The equation of projectile motion is given by :


h(t)=-16t^2+ut+h_o

According to the given condition,


h(t)=-16t^2+36t+4

h(t) = 2.356 meters

Height of the ball at t = 2 s


h(2)=-16(2)^2+36(2)+4

h = 12 ft

So, at t = 2 seconds the height of the ball is 12 feet from the ground. Hence, this is the required solution.

User Mahmoud Ali
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