Question

The maximum height attained by a shell is 80 m & the horizontal range is 320m what is the velocity & angle of projection

Answer 1

Answer:

angle of projection=45° and velocity=56.56m/s

Explanation:

H=u^3sin^2teta/2g.............(1)

Range=u^2sin2teta/g.........(2)

when u divide these two equation u will be able to get angle of projection,

H/R=u^2sin^2teta/2g / u^2sin2teta/g

H/R=u^2sin^2teta/2g× g/u^2sin2teta

H/R=sin^2teta/2sin2teta

height=80m

range=320m

when u apply them,

80/320=sin^2teta/2sin2teta

1/4=sin^2teta/2sin2teta

1/²=sinteta/2sintetacosteta ( sin2teta=2sintetacosteta)

1=sinteta/costeta

1=tanteta

teta= 45°

when u apply this velocity to any of the equation in(1)or(2) u will get velocity,

R=u^2sin2teta/g

320=u^2sin2(45)/g

320=u^2sin90/g

320=u^2(1)/g

320=u^2(1)/10

3200=u^2(1)

√3200=u

u=56.56m/s