# Child sitting in a tree throws his apple from the height of 4.0 m with a velocity of 5.0 m/s [35 degrees above horizontal]. It hits the ground right next to his friend. How long is it before the...

Child sitting in a tree throws his apple from the height of 4.0 m with a velocity of 5.0 m/s [35 degrees above horizontal]. It hits the ground right next to his friend. How long is it before the apple core hits the ground?

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### 1 Answer

For this question, you can use:

`dy= V_(iy) * t + 1/2at^2`

Now first we shall assign the values needed to solve for the time required.

`dy` = -4 (direction is down)

`V_iy` = 5m * sin (35) = 2.8679

`a ` = acceleration due to gravity = -9.81m/s^2 (direction is down)

`t ` = time = ?

By substitution, we can now have:

`-4 = 2.8679t - 4.9t^2`

Arranging the equation into standard form:

`4.9x^2 -2.8679x - 4 = 0`

To solve for the value of t, quadratic equation is used.

`x = (-b +-sqrt(b^2 - 4ac))/(2a)`

The two values derived from this equation are -0.657 and 1.24. we will choose the positive time which is 1.24 seconds.

**Final answer: 1.24 ~ 1.2 seconds**

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