# Question 04fd1

Jan 24, 2017

The ball will reach a max height of $\underline{45}$ m in $\underline{3}$ seconds

#### Explanation:

We use the kinematic equation connecting various quantities of interest. The maximum height is reached when upwards velocity is $= 0$
1. To calculate Max height $h$ the kinematic equation is
${v}^{2} - {u}^{2} = 2 a h$
Final velocity $v$ after time $t$, $a$ is acceleration and initial velocity is $u$

Inserting given values we get, remember, direction of acceleration due to gravity is opposite to the direction of initial velocity

${0}^{2} - {30}^{2} = 2 \times \left(- 10\right) h$
$\implies 20 h = 900$
$\implies h = \frac{900}{20} = 45 m$
2. To calculate time taken to reach maximum height we use the other kinematic equation
$v = u + a t$

Inserting given values we get

$0 = 30 - 10 \times t$
$\implies 10 \times t = 30$
$\implies t = \frac{30}{10} = 3 s$

Jan 24, 2017

h= 45 m; t = 3 s

#### Explanation:

The initial Kinetic energy (where potential energy is zero) becomes only potential energy (where the ball reaches the max height and its speed is zero).
$\frac{1}{2} m {v}^{2} = m g h$
but, since the mass is the same
$\frac{1}{2} {v}^{2} = g h$
and
$h = {v}^{2} / \left(2 g\right)$
$h = \frac{900 {m}^{2} / {s}^{2}}{2 10 \frac{m}{s} ^ 2} = 45 m$
from the second law of motion
v= v°-at
$0 = 30 \frac{m}{s} - 10 \frac{m}{s} ^ 2 t$
$t = 3 s$
now let's verify in the first law of motion
s= v°t -1/2 g t^2#
$45 m = 30 \frac{m}{s} 3 s - \frac{1}{2} 10 \frac{m}{s} ^ 2 {\left(3 s\right)}^{2}$
45 = 45 m