Question

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it falls back to the ground. a. What height is the object thrown from? b. How long does it take for the object to hit the ground?

Answer 1

The value is
`T_t = 2.5659 \ s`

Step-by-step explanation:

From the we are told that

The initial speed of the object is
`u = 8 \ m/s`

The greatest height it reached is
`h = 15 \ m`

Generally from kinematic equation we have that

`v^2 = u^2 + 2gH`

At maximum height v = 0 m/s

So

`0^2 = 8^2 + 2 * - 9.8 * H`

=>
`H = 3.27 \ m`

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

`s = h - H`

=>
`s = 15 - 3.27`

=>
`s = 11.73 \ m`

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

`v = u + (-g) t`

At maximum height v = 0 m/s

`0 = 8 - 9.8t`

=>
`t = 0.8163 \ s`

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

`h = ut_1 + (1)/(2) gt_1^2`

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

So

`15 = 0* t + 4.9 t^2`

=>
`t_1 = 1.7496`

Generally the total time taken is mathematically represented as

`T_t = t + t_1`

=>
`T_t = 0.8163 + 1.7496`

=>
`T_t = 2.5659 \ s`