Since the initial velocity is 34.3 m/s with initial height of 2.1 m, then upon substituting into the height formula, we get:

`h=-1/2(9.8)t^2+34.3t+2.1`

The corresponding velocity formula for constant acceleration is

`v=-g t+v_0`

And the maximum height is found when v=0. This gets:

`0=-9.8 t+34.3` solve for t

`9.8t=34.3` divide...

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Since the initial velocity is 34.3 m/s with initial height of 2.1 m, then upon substituting into the height formula, we get:

`h=-1/2(9.8)t^2+34.3t+2.1`

The corresponding velocity formula for constant acceleration is

`v=-g t+v_0`

And the maximum height is found when v=0. This gets:

`0=-9.8 t+34.3` solve for t

`9.8t=34.3` divide by 9.81

`t=34.3/9.8=3.5`

The object reaches maximum height at 3.5 seconds after launch. This is now substituted into the height formula to get:

`h=-4.9(3.5)^2+34.3(3.5)+2.1=62.125`

**The maximum height is 62.1m and the time it occurs is at 3.5 s.**