Question

An object is projected so that it just clears two wall of height 7.5 m and with seperation 50 m from each other . If the time passing between the walls is 2.5 s, the range of the projectile will be: $(g=10m/s^2)$

• A. 35 m
• B. 70 m
• C. 140 m
• D. 57.5 m

Answer 1

The correct option is B 70 m

Horizontal component of projectile's motion

$\frac{50}{2.5}=20㎧$

To have a range of $50m$, maximum height would be reached in $\frac{2.5}{2}=1.25secs$ (neglecting walls for now)

$1.25\times g=$initial velocity$=12.5㎧$

Height attained over walls$=\frac{V^2}{2g}$

$=\frac{{\left(12.5\right)}^2}{2g}=7.8125m$

Adding wall weight$(7.812+7.5)=15.3125m$

above ground level

So initial vertical v component from ground level

$=\sqrt{2gh}$

$=\sqrt{2\times 10\times 15.3125}=17.5㎧$

Time to maximum height from the ground$=\frac{V}{g}$

$=\frac{17.5}{10}=1.75sec$

Total time in air$=(1.75\times 2)=3.5sec$

Range$=3.5\times 20$

$=70m$