# A projectile is launched upward from ground level with an intentional speed of 98 m/s. How high will it go? when will it return to the ground?

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Expert Answers

jerichorayel | Certified Educator

The projectile motion can be represented as:

`vf^(2) - vi^(2) = -2gy` and

`vf = vi -g t` where

vf = final velocity

vi = initial velocity = 98m/s

g = acceleration due to gravity = `9.8 (m)/(s^(2))`

y = distance in meters

At the highest peak, the velocity is zero. Therefore:

`vf^(2) - vi^(2) = -2gy`

`0 - 98(m)/(s^2) = -2(9.8(m)/(s^2))y`

`-9604(m^2)/(s^2) = (-19.6(m)/(s^2))y`

`y = (-9604(m^2)/(s^2))/(-19.6(m)/(s^2))`

**y = 490 m -> distance at the highest peak**

To get the time it will fall:

`vf = vi -g t`

`0 = 98m/s -(9.8(m)/(s^2))* t`

`t = (98m/s)/(9.8 (m)/(s^2))`

**t = 10 seconds**