Question

a particle is projected at an 30°from the horizontal having initial velocity 72km/h . find the horizontal range time of flight and max height attend by the particle​

Answer 1

Answer:

Time of flight =

g

2usinθ

=

10

2×80×sin30

o

=8

As particle during journey at t=2sec and at t=6sec will be at same height and hence vertical displacement is zero.

Horizontal displacement =(ucosθ)×(6−2)

=80×

2

3

×4

=160

3

m

∴ Average velocity =

4

160

3

=40

3

m/sec

Answer 2

Explanation:

time of flight =

g

2usin0

=

10

2×80×sin 30

0

=8

as particle during journey at t=2sec and at t=6sec will be same height and hence vertical displacement is zero.

horixontal displacement ={ucos0}×{6-2}

=80×

2

3

×4

=160

m

average velocity=

4

160

3

=40

3

m/sec