# the formula for height, h meterof an object propelled into the  air is h=-1/2gt^2 + v0t + h0 where 'g' is the acceleration due to gravity is 9.8m/s^2, 't' represents the time in seconds, 'v0' represents the initial velocity as m/s, 'h0' represents the initial height in meters. A projectile has an initial velocity 34.3 m/s^2 and is launched 2.1 meter above the ground. 1. find the maximum height in meters reached by the projectile? 2. How many seconds after the launch does the projectle reach the maximum height

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.

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