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3 years ago

5

A ball is launched into the sky at 48 feet per second from a building 48 feet tall. The equation for the ball’s height h at time

t seconds is h = -16t2 + 48t + 48. When will the ball strike the ground?

1 answer:

kkurt [141]3 years ago

3 0

Answer:

4 seconds I believe

Step-by-step explanation:

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Sergeeva-Olga [200]

Answer:

(C) $6t^2+5$

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

$d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.$

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

$v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5$

The correct option is C.

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