# An object is thrown upward with a speed of 8m/s from the roof of a building 10m high. It rises and then falls back until it strikes the ground. What is the greatest height above the ground that the object attains?With what speed does object strike ground?

## This is what I got so far: $a \left(t\right) = \frac{\mathrm{dv}}{\mathrm{dt}} = - 9.8 \frac{m}{s} ^ 2$ $v \left(t\right) = - 9.8 t + C$ (Antideriv of a(t) $v \left(0\right) = 8 , 8 = 0 + c \to C = 8$ $v \left(t\right) = - 9.8 t + 8$ $s \left(t\right) = - 4.9 {t}^{2} + 8 t + C$ $s \left(0\right) = 10 , 10 = 0 + 0 + C \to C = 10$ $s \left(t\right) = - 4.9 {t}^{2} + 8 t + 10$ $s \left(t\right) = 0$ $- 4.9 {t}^{2} + 8 t + 10 = 0$ $t = \frac{- 8 \left(\frac{+}{-}\right) \sqrt{260}}{9.8}$ $t = 8 + \frac{\sqrt{260}}{9.8} = 2.46 \ldots$ (cant be neg) I have no idea what to input for a.. For b, however, I think that the speed when the object strikes the ground is $v = a t = - 9.8 \left(2.46 \ldots\right) = - 24.125 \frac{m}{s}$

Dec 1, 2017

Initial upward velocity of the object from the roof of the 10m high building $u = + 8 m \text{/} s$ (taking upward direction as +ve)

So downward acceleration due to gravity $g = - 10 m \text{/} {s}^{2}$

At its maximum height $h$ from the roof its velocity will become zero.

So ${0}^{2} = {u}^{2} - 2 \times g \times h$

$\implies h = {u}^{2} / \left(2 g\right) = {8}^{2} / \left(2 \cdot 10\right) = 3.2 m$

Hence the maximum height attained by the object from the ground will be

$H = h + 10 = 3.2 + 10 = 13.2 m$

When the object reaches the ground its net downward displacement becomes $d = - 10 m$ . If the object strikes the ground with speed $v m \text{/} s$
Then we can write

${v}^{2} = {u}^{2} + 2 g d$

$\implies {v}^{2} = {8}^{2} + 2 \left(- 10\right) \cdot \left(- 10\right) = 264$

$\implies v = \sqrt{264} \approx 16.24 m \text{/} s$