A body is projected with a velocity of 40 m/s. After 2sec if crosses a vertical pole of height 20cm. Calculate the angle of projection and horizontal range.

Open in App

Solution

$x = 40 cos θ \times 2 = 80 cos θ$
Applying equation of trajectory ,
$20 = 80 cos θ \times tan θ - \frac{g (80 cos θ)^{2}}{2 \times {cos}^{2} θ \times 40^{2}}$
$20 = 80 sin θ - \frac{40}{2}$
$sin θ = \frac{1}{2}, θ = 30^{\circ}$
Range $= \frac{u^{2} sin 2 θ}{g} = \frac{1600 \times \sqrt{3}}{20} = 80 \sqrt{3} = 138.5 m$

Suggest Corrections

Similar questions

A body is projected with a velocity of 40ms^-1.After 2s,it crosses a vertical pole of length 20 calculate the angel of projection and horizontal range, g=10 ms^-2

Q. A body is projected with velocity

u

such that its horizontal range and maximum vertical heights are same. The maximum heights is:

Q. A body is projected with a velocity of 20 m/s in a direction making an angle of

60^{\circ}

with the horizontal. Calculate
(i) its position after 0.55
(ii) velocity after 0.55

Q. A body is projected with an initial velocity of 50

m s^{- 1}

at angle 53 with the horizontal. Calculate the velocity when it reaches a height of 40 m.

Q. A body, projected vertically up with a velocity of

20 m s^{- 1}

, reaches a height of

20 m

. If it is projected with a velocity of

40 m s^{- 1}

, then the maximum height reached by the body is:

Join BYJU'S Learning Program

A body is projected with a velocity of 40 m/s. After 2sec if crosses a vertical pole of height 20cm. Calculate the angle of projection and horizontal range.

x=40cosθ×2=80cosθApplying equation of trajectory ,20=80cosθ×tanθ−g(80cosθ)22×cos2θ×40220=80sinθ−402sinθ=12,θ=30∘Range =u2sin2θg=1600×√320=80√3=138.5m