A ball is thrown upward from the top of a 200 foot tall building with a velocity

Question

A ball is thrown upward from the top of a 200 foot tall building with a velocity of 40 feet per second. Take the positive direction upward and the origin of the coordinate system at ground level. What is the initial value problem for the position, x(t), of the ball at time t? Select the correct answer. If you could please explain how to obtain the correct answer, I would appreciate it. Thanks!

a) d2x/dt2 = 40 , x(0) = 200 , dx/dt(0) = 40

b) d2x/dt2 = -40 , x(0) = 200 , dx/dt(0) = 40

c) d2x/dt2 = 32 , x(0) = 200 , dx/dt(0) = 40

d) d2x/dt2 = 200 , x(0) = 32 , dx/dt(0) = 40

Explanation / Answer

c) d2x/dt2 = 32 , x(0) = 200 , dx/dt(0) = 40

because acceleration is 9.8 m/sec^2

that is 32.15 feet/sec^2

dx^2/dt^2=32 feet/sec^2

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