A ball is dropped from the top of teh space needle to the ground below. The heig

Question

A ball is dropped from the top of teh space needle to the ground below. The height of the ball above tha ground in feet is a function of time in seconds given by y=605-16t^2

(a) find the velocity fo the ball at time t. What is the sign of teh velocity(neg. or pos.). Why dies this make sense

(b) waht is th eacceleration of teh ball? It shoiuld be constant-Why?

(c) when does the ball hit the ground and how fast os ot going at teh time? Give your anser in feet per seconds and in miles per hour (1ft/sec = 15/22mph).

Explanation / Answer

a) v = dy/dt = -32*t

negative which makes sense since its falling down

b) a = dv/dt = -32

constant since falling due to gravity

c) hits the ground when y = 0

605- 16 t^2 = 0

t = sqrt(605/16)=6.15 s

v = -32*6.15=-196.8 ft/s = 134.2 mph

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