Question: In physics, it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from ground level

In physics, it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from ground level is  

s = s(t) = - 16t2 + 96t

where t is the elapsed time that the ball is in the air. 

(a) When does the ball strike the ground? That is, how long is the ball in the air? 

(b) What is the average speed of the ball from t = 0 to t = 2?

(c) What is the instantaneous speed of the ball at time t? 

(d) What is the instantaneous speed of the ball at t = 2?

(e) When is the instantaneous speed of the ball equal to zero? 

(f) How high is the ball when its instantaneous speed equals zero? 

(g) What is the instantaneous speed of the ball when it strikes the ground?

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a b 16t 96t0 16tt60 t0 or t 6 The ball strikes the ground after 6 seconds Ass2s0 20 At 162 9620 ... View full answer

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