Question

A ball is thrown with a velocity whose horizontal component is $12{\text{ms}}^{-1}$ from a point $15\text{m}$ above the ground and $6\text{m}$ away from a vertical wall $18.75\text{m}$ high is such a way to just clear the wall. At what time(in $\text{s}$)will be ball reach the ground? (take $g=10{\text{ms}}^{-2})$

Answer 1

Horizontal component of velocity, $u\cos \theta =12,$
Using 2nd equation of motion in horizontal and vertical direction respectively when ball comes to same horizontal level from where its projection started i.e. from A to B,
$\begin{array}{}6=u\cos \theta t\Rightarrow t=0.5\text{s}& \text{(1)}\end{array}$
$\begin{array}{}3.75=u\sin \theta t-\frac{1}{2}g{t}^2& \text{(2)}\end{array}$

Putting the value of $t=0.5s$ in $\left(2\right)$,
$\Rightarrow \frac{15}{4}=u\sin \theta \times \frac{1}{2}-\frac{1}{2}10{\left(\frac{1}{2}\right)}^2$
$\Rightarrow u\sin \theta =10$
Using 2nd equaton of motion in vertical direction from B to C,
$-15=10T-\frac{1}{2}10T^2\Rightarrow T^2-2T-3=0T=3,-1\Rightarrow T=3\text{s}$