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A ball is thrown straight up in the air with an initial velocity of 50 feet per second and an initial height of 5 feet. How high will the ball be after 3 seconds?
Wiki User
∙ 13y ago
Step 1
Write down the information you already know:
v1= 50 feet per second [up]
a= 9.81 m/s2 [down]= -9.81 m/s2 [up] = -32.185 feet/s2 [down] (1 metre= 3.28083989501312 feet)
t= 3s
d= ?
Step 2
Solve:
d= v1t+(1/2)at2
d= 50(3)+(1/2)(-32.185)(3)2
d= 150-144.8325
d= 5.1675 feet
*Remember you need to add the initial height
d=5+5.1675
d= 10.1675 feet= 10 feet
Wiki User
∙ 13y ago
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