Question

Two stones A and B are projected simultaneously as shown in figure. It has been observed that both the stones reach the ground at the same place after $7$ seconds of their projection. The difference in the vertical components of their initial velocities in $\text{m/s}$ is ($g=9.8{\text{m/s}}^2$)

Answer 1

For ball B:
Using second equation of motion, we can say
$-49=u_2\sin {\theta }_{2}t-\frac{1}{2}g{t}^2$
On putting $t=7\text{s}$ in the above equation, we get
$\Rightarrow u_2\sin {\theta }_2=27.3\text{m/s}$

For Ball A:
Using second equation of motion, we can say
$0=u_1\sin {\theta }_{1}t-\frac{1}{2}g{t}^2$
On putting $t=7\text{s}$ in the above equation, we get
$\Rightarrow u_1\sin {\theta }_1=34.3\text{m/s}$
$\therefore$ Difference in vertical components of the initial velocities of balls $=u_1\sin {\theta }_1-u_2\sin {\theta }_2=34.3-27.3=7\text{m/s}$