# A body thrown vertically upward returns to the earth in three seconds.(a)what was the initial velocity of the body (b)what height did the body reach? Neglect air resistance.

Aug 31, 2017

a 14.7 m/s
b 11 meters

#### Explanation:

A couple of facts we know. Gravitational acceleration in the vicinity of the earth's surface can be treated as a constant 9.81 m/s.

This is $\frac{{d}^{2} x}{\mathrm{dt}} ^ 2$

With this, we can write a function for velocity v as a funciton of t:

$v \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = \int - 9.81 \mathrm{dt} = {v}_{0} - 9.81 t$

(We are assigning a NEGATIVE value to downward acceleration & velocity)

We are given total flight time: 3 seconds.

So we know that at the halfway point (1.5 seconds), the ball will have reached maximum height, and v at that time will be 0.

The ball then accelerates downwards for the remaining 1.5 seconds.

Over this time it accelerates to $- 9.81 \cdot 1.5 = - 14.7 \frac{m}{s}$.

Since we're ignoring wind resistance, the downwards velocity at time t = 3 is equal in magnitude (opposite in direction) to the initial velocity. So this is your answer to part a: 14.7 m/s.

(I'm rounding to 3 significant digits in all arithmetic calculations).

We can then write an equation for the height, x, as a function of t:

$x \left(t\right) = \int \left({v}_{0} - 9.81 t\right) \mathrm{dt} = \left({v}_{0} \cdot t\right) - \left(\frac{9.81}{2}\right) {t}^{2} + c$

At time t = 0, height x = 0, so our constant c is also 0.

Plugging in time t = 1.5, we can solve for the height at that time:

$x \left(1.5\right) = 14.7 \left(1.5\right) - 4.91 \left({1.5}^{2}\right) = 11 m e t e r s$

(once again, rounding to 3 significant digits.)