# Question 4f1eb

Feb 24, 2017

Explanation:

#### Explanation: We know the ball is thrown upwards in a initial velovity of $15.8 \frac{m}{s}$

So, $u = 15.8 \frac{m}{s}$

When the ball reaches its maximum height, it has a velocity of $0 \frac{m}{s}$

So, $v = 0 \frac{m}{s}$

Let the maximum height be $x$

The gravitational force that makes the ball slow down till it reaches its maximum height

So, $a = - 9.8 \frac{m}{s}$

Now, we have to find the time to reach its maximum height

We have our values

color(orange)(u=15.8 m/s
color(orange)(v=0 m/s
color(orange)(a=-9.8 m/s

We can use one of the Kinematic equations

color(blue)(v=u+at

$\rightarrow 0 = 15.8 + \left(- 9.8 t\right)$

$\rightarrow 0 = 15.8 - 9.8 t$

$\rightarrow 9.8 t = 15.8$

$\rightarrow t = \frac{15.8}{9.8}$

color(green)(rArrt=1.61 s

We can use this value to find the maximum height ($x$) using another Kinematic equation

color(brown)(x=ut+1/2at^2

$\rightarrow x = \left(0 \cdot 1.61\right) + \left(\frac{1}{2} \cdot \left(- 9.8\right) \cdot {1.61}^{2}\right)$

$\rightarrow x = \frac{1}{2} \cdot \left(- 9.8\right) \cdot 2.59$

$\rightarrow x = \left(- 9.8\right) \cdot 1.29$

color(green)(rArrx=-12.64m#

As height cannot be negative, $x = 12.64 m$

Hope this helps! :)