Ball thrown from top of building

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A rock is thrown from a 100-ft tall building and lands 160 ft away after 5.0 seconds. The horizontal component of the initial velocity is calculated to be 32 ft/s. To determine the initial vertical velocity, the vertical displacement at the time of impact must be used rather than focusing on maximum height. The relevant equations involve the initial vertical velocity and gravitational acceleration.

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rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

I found the x component of the initial velocity to be 32m/s and I think I need to find the y(max). I found t(max) for height to be v(0)/g (for the y component). And y(tmax)=h+1/2(v(o)/g), but this information does not seem to help me any.

Help?
 
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Hi jhoffma4,

jhoffma4 said:
rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

I found the x component of the initial velocity to be 32m/s

Be careful with your units; this will be 32 ft/s.

and I think I need to find the y(max). I found t(max) for height to be v(0)/g (for the y component). And y(tmax)=h+1/2(v(o)/g)

This equation does not look correct to me.

, but this information does not seem to help me any.

Help?

I would not worry about the maximum height. You know the vertical displacement at a particular time after launch, right? That's all you need to find the initial vertical velocity. What do you get?
 

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