# Question b18b1

Dec 24, 2017

#### Explanation:

The initial velocity given in the question is,${v}_{0} = 125 m {s}^{- 1}$

The throwing angle, theta_0=20°

The mentioned height from ground,$y = 58.6 m$

We know that,

color(green)(y=v_0sintheta_0t-1/2g(t)^2
:.56.8=125sin20°t-1/2 9.8(t)^2
$\mathmr{and} , 56.8 = 42.75 t - 4.9 {\left(t\right)}^{2}$

$\mathmr{and} , 4.9 {\left(t\right)}^{2} - 42.75 t + 56.8 = 0$

$\therefore t = \frac{- \left(- 42.75\right) \pm \sqrt{{42.75}^{2} - 4 \times 4.9 \times 56.8}}{2 \times 4.9}$
$= \frac{42.75 \pm 26.73}{9.8} = 1.64 s \mathmr{and} 7.09 s$
$W h e n , t = 1.64 s$,then the distance from the throwing point to the hitting point of the projectile,

x=v_0costheta_0t=125ms^(-1)xxcos20°xx1.64s=192.7m#

$W h e n , t = 7.09 s$,then the distance from the throwing point to the hitting point of the projectile,$x = 833.08 m .$

So,The building will be$192.7 m \mathmr{and} 833.08 m$ far from the throwing point$\textcolor{b r o w n}{\left(A n s . \left(a\right)\right)}$

As a projectile moves in semi- circular way, its height from the ground would same for two different times$\textcolor{\mathmr{and} a n \ge}{\left(A n s . \left(b\right)\right)}$

You also want the tallest building that the projectile could clear. It means the highest height$\textcolor{red}{' H '}$ of the projectile from the ground

we know,$\textcolor{b l u e}{H = \frac{{v}_{0}^{2} {\sin}^{2} {\theta}_{0}}{2 g}}$$= 92.156 m \textcolor{red}{\left(A n s . \left(c\right)\right)}$