Question

A ball is thrown at 14.7 m/s from the top of a 88.2 meter tall building. The equation for the balls height s at time t seconds after the ball is thrown; s(t) = –4.9t2 + 14.7t + 88.2. How many seconds does the ball take to hit the ground?
A) 2
B) 3
C) 4
D) 6

Answer 1

Answer:

The ball would take t = 6 seconds to hit the ground. So, option (D) is correct.  

Step-by-step explanation:

As the equation for the balls height s at time t seconds after the

ball is thrown;

                       s(t) = –4.9t² + 14.7t + 88.2

When a ball hits the ground, the height (s) would be zero.

So,

                      0 = –4.9t² + 14.7t + 88.2

                      4.9t² − 14.7t − 88.2 = 0

Use quadratic formula with a = 4.9, b = -14.7, c = -88.2.

t =  −b ±√b²−4ac ÷  2a

t = −(−14.7) ±√(−14.7)²−4(4.9)(−88.2)  ÷  2(4.9)

t =  14.7±√1944.81  ÷ 9.8

t = 6, −3        (ignore the negative value)

t = 6 seconds

Hence, the ball would take t = 6 seconds to hit the ground. So, option (D) is correct.  

Keywords: time, second, height s at time t

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