Question

An object is projected so that it just clears two walls of height 7.5 m and with separation 50 m from each other. If the time passing between the walls is 2.5 s, the range of the projectile will be ..... (g = 10 m/s^2
a) 35 m
B) 70 m
C) 140 m
D) 57.5 m

Answer 1

Horizontal component of projectile's motion = (50/2.5) = 20m/sec.
To have a range of 50m., maximum height would be reached in (2.5/2) = 1.25 secs. (neglect walls for now).
(1.25 x g) = initial vertical V of 12.5m/sec.
Height attained over walls = (v^2/2g) = (12.5^2/20) = 7.8125 metres. Now add in the wall height.
(7.8125 + 7.5) = 15.3125 metres above ground level.
So, initial vertical V component from ground level = sqrt.(2gh) = sqrt. (20 x 15.3125) = 17.5m/sec.
Time to max. height from ground = (v/g) = 17.5/10, = 1.75 secs.
Total time in air = (1.75 x 2) = 3.5 secs.
Range = (3.5 x 20m/sec.) = 70 metres.