A projectile is fixed at an angle $60^{\circ}$ with horizontal. Ratio of initial K.E. to K.E. when velocity vector of projectile makes an angle $15^{\circ}$ with velocity of projection is

Open in App

Solution

v = $\frac{u c o s 60^{\circ}}{c o s 45^{\circ}}$

$∴ K_{f} = K_{i} \frac{c o s^{2} 60^{\circ}}{c o s^{2} 45^{\circ}}$

$∴ \frac{K_{i}}{K_{f}} = 2$

Suggest Corrections

Similar questions

Q. A body is projected such that its

K . E .

at the top is

\frac{3}{4}

of its initial

K . E .

. What is the angle of projectile with the horizontal?

Q. A particle is projected from the ground at an angle of

θ

with the horizontal with an initial speed of

u

. Time after which velocity vector of the projectile is perpendicular to the initial velocity is

Q. A projectile is projected with a velocity of

v_{0}

at an angle

θ

with the horizontal. What is the ratio of the rate of change of speed with respect to time at the point of projection to that at a point where velocity makes an angle

θ / 2

A projectile is projected with velocity of 50m/s at an angle of 60^° with the horizontal from the ground. The time after which is velocity will make an angle of 45^°with the horizontal is

1)2.5s

2)1.83s

3)2.37s

4)3.72s

Q. If a projectile is projected an angle

30^{\circ}

with respect to the horizontal with an initial velocity 4 m/s, range of projectile is

Join BYJU'S Learning Program

A projectile is fixed at an angle 60∘ with horizontal. Ratio of initial K.E. to K.E. when velocity vector of projectile makes an angle 15∘ with velocity of projection is

v = ucos 60∘cos 45∘ ∴Kf=Kicos2 60∘cos2 45∘ ∴KiKf=2

A projectile is fixed at an angle $60^{\circ}$ with horizontal. Ratio of initial K.E. to K.E. when velocity vector of projectile makes an angle $15^{\circ}$ with velocity of projection is

v = $\frac{u c o s 60^{\circ}}{c o s 45^{\circ}}$

$∴ K_{f} = K_{i} \frac{c o s^{2} 60^{\circ}}{c o s^{2} 45^{\circ}}$

$∴ \frac{K_{i}}{K_{f}} = 2$