Question

A ball is thrown upward at an angle of ${30}^{\circ }$ with the horizontal and lands on the top edge of a building that is $20m$ away. The top edge is $5m$ above the throwing point. The initial speed of the ball in $metre/second$ is (take g = 10 $m/s^2$)

• A. $15m{s}^{-1}$
• B. $20m{s}^{-1}$
• C. $25m{s}^{-1}$
• D. $30m{s}^{-1}$

Answer 1

Answer: (A), (B)

The correct options are
A $15m{s}^{-1}$
B $20m{s}^{-1}$

$20m/s$
The key problem like this is to treat the horizontal and vertical components of the motion separately and use the fact that they share the same time of flight.
Horizontal component
The horizontal component of the velocity is constant as it is perpendicular to $g$.
So we can write:
$v\cos 30=\frac{20}{t}$
Where $t$ is the time of flight
$\therefore v\times 0.866=\frac{20}{t}$
$t=\frac{20}{0.866v}$ $\left(1\right)$
Vertical Component
We can use :
$s=ut+\frac{1}{2}a{t}^2$
This becomes :
$5=v\sin 30t-\frac{1}{2}g{t}^2$
$\therefore 5=v\times 0.5\times t-\frac{1}{2}\times 9.8\times t^2$
$5=v\times 0.5\times t-4.9t^2\left(2\right)$
We can substitute the value of $t$ from $\left(1\right)$ into the first part of $\left(2\right)\Rightarrow$
$\therefore 5=\frac{v\times 0.5\times 20}{0.866v}-4.9t^2$
$\therefore 5=\frac{10}{0.866}-4.9t^2$
$4.9t^2=11.547-5=6.547$
$4.9t^2=6.547$
$t^2=\frac{6.547}{4.9}=1.336$
$t=\sqrt{1.336}=1.156s$
We can now put this value of $t$ back into $\left(1\right)\Rightarrow$
$\frac{20}{0.866v}=t$
$\therefore 0.866v=\frac{20}{t}=\frac{20}{1.156}$
$\therefore v=\frac{20}{1.156\times 0.866}=19.97m/s$