Question

A projectile crosses two walls of equal height H symmetrically. The time of flight is

Answer 1

Answer:

let h be the height, u be the initial velocity and t be the time taken .

Now h=ut−

2

gt

2

At t=2s and t=6s the heights are equal

∴2u−

2

2

2

g

=6u−

2

6

2

g

4u=16g

u=40m/s

Height of each wall,

h=ut−

2

gt

2

=80−20=60m

Time of flight =

g

2u

=

10

80

=8s

Maximum height of projectile =

2g

u

2

=80m