Question

Two adjacent bowling machines deliver balls at 30.5 m/s simultaneously at angles of elevation ${\alpha }_1\text{and}{\alpha }_2$ for the same target at a range of 20.13 m. Calculate the time difference between the two hits

• A. 5.5 sec
• B. 55 sec
• C. 11 sec
• D. 22 sec

Answer 1

The correct option is A 5.5 sec

Let ${\alpha }_{1}and{\alpha }_2$ be the angles of elevation

$\text{Horizontal range,}R=\frac{v_0^2}{g}sin2\alpha$

Since the range is the same

$\frac{v_0^2}{g}sin2{\alpha }_1=\frac{v_0^2}{g}sin2{\alpha }_2$

$sin2{\alpha }_1=sin2{\alpha }_2=sin(180-2a_2)$

$\Rightarrow {\alpha }_1=90-{\alpha }_2$

i.e., the projectile will have the same range for two angles ${\alpha }_1\&{\alpha }_2$ such that ${\alpha }_1+{\alpha }_2={90}^o$, provided $v_0$ is the same.

$i.e.,20.13=\frac{(30.5{)}^2}{9.81}\times sin2{\alpha }_1$

$\therefore {\alpha }_1={6.21}^0$

$\therefore {\alpha }_2={83.88}^0$

$\therefore \text{Time of flight}{t}_1=\frac{2v_o}{g}sin{\alpha }_1$

$t_2=\frac{2v_o}{g}sin{\alpha }_2$

Time Difference between hits

$t_2-t_1=\frac{2v_0}{g}[sin{\alpha }_2-sin{\alpha }_1]=5.52sec$