A projectile of 2 kg has velocities 3 m/s and 4 m/s at two points during its flight in the uniform gravitational field of the earth…

Q: A projectile of 2 kg has velocities 3 m/s and 4 m/s at two points during its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other then the minimum KE of the particle during its flight is

Sol. v1 cos⁡α = v2 cos⁡(90-α)

3 cos α = 4 sin α

$\large tan\alpha = \frac{3}{4} $

$\large K.E_{min} = \frac{1}{2}m v_1^2 cos^2\alpha$

$\large = \frac{1}{2}\times 2 \times 3^2 (\frac{4}{5})^2 $

= 5.76 J

A foot ball is kicked off with an initial speed of 19.6 m/s to have maximum range. Goal keeper standing on the goal line…

Q: A foot ball is kicked off with an initial speed of 19.6 m/s to have maximum range. Goal keeper standing on the goal line 67.4 m away in the direction of the kick starts running opposite to the direction of kick to meet the ball at that instant. What must his speed be if he is to catch the ball before it hits the ground ?

Sol: $\large R = \frac{u^2 sin2\theta}{g}$

$\large R = \frac{(19.6)^2 sin90^o}{9.8}$

R = 39.2 meter.

Man must run 67.4 m – 39.2 m = 28.2 m in the time taken by the ball to come to ground Time taken by the ball.

$\large t = \frac{2 u sin\theta}{g} $

$\large t = \frac{2 \times 19.6 sin45^o}{9.8} = \frac{4}{\sqrt{2}}$

t = 2√2 = 2 × 1.41 = 2.82 sec.

Velocity of man $\large = \frac{28.2}{2.82} $

= 10 m/s

The velocity of a projectile at its greatest height is √(2/5) times its velocity, at half of its greatest height…

Q: The velocity of a projectile at its greatest height is √(2/5) times its velocity, at half of its greatest height, find the angle of projection.

Sol: $\large u cos\theta = \sqrt{\frac{2}{5}} \times u \sqrt{\frac{1+cos^2\theta}{2}}$

Squaring on both sides

$\large u^2 cos^2 \theta = \frac{2}{5}u^2 (\frac{1+cos^2 \theta}{2}) $

$\large 4 cos^2 \theta = 1 $

$\large  cos^2 \theta = \frac{1}{4}$

θ = 60°

A body is projected with velocity u at an angle of projection θ with the horizontal. The body makes 30° with horizontal…

Q: A body is projected with velocity u at an angle of projection θ with the horizontal. The body makes 30° with horizontal at t = 2 second and then after 1 second it reaches the maximum height. Then find
(a) Angle of projection
(b) Speed of projection.

Sol. During the projectile motion, angle at any instant t is

$\Large tan\alpha = \frac{v_y}{v_x} = \frac{usin\theta – g t}{u cos\theta}$

For t = 2 seconds, α =30°

$\Large \frac{1}{\sqrt{3}} = \frac{usin\theta – 2g}{u cos\theta}$ ….(i)

For t = 3 seconds, at the highest point α = 0°

$\Large 0 = \frac{usin\theta – 3g}{u cos\theta}$ ….(ii)

Solving (i) & (ii)

θ = 60° , u = 20√3 m/s

The ceiling of a lone hall is 20 m high. What is the maximum horizontal distance that a ball thrown with…

Q: The ceiling of a lone hall is 20 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall (g = 10 ms2 ) ?

Sol. Here, H = 20 m, u = 40 m/s

Suppose the ball is thrown at an angle θ with the horizontal.

$\large H = \frac{u^2 sin^2 \theta}{2 g} $

$\large 20 = \frac{(40)^2 sin^2 \theta}{2 \times 10} $

sinθ = 1/2 ⇒ θ = 30°

$\large R = \frac{u^2 sin2\theta}{g} $

$\large R = \frac{(40)^2 sin60^o}{10} $

= 138.56 m

A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection…

Q: A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Sol: Horizontal Range , $\large R = \frac{u^2 sin2\theta}{g} $

$\large 3 = \frac{u^2 sin60^o}{g} $

$\large 3 = \frac{u^2}{g} \times \frac{\sqrt{3}}{2} $

$\large \frac{u^2}{g} = 2 \sqrt{3} km$ …(i)

$\large R_{max} = \frac{u^2}{g} $

$\large R_{max} = 2 \sqrt{3} = 2 \times 1.732 = 3.464 km$

So, it is not possible to hit the target 5km away.

A swimmer crosses a flowing stream of width ‘d’ to and fro normal to the flow of the river in time t1…

Q: A swimmer crosses a flowing stream of width ‘d’ to and fro normal to the flow of the river in time t1. The time taken to cover the same distance up and down the stream is t2 . If t3 is the time the swimmer would take to swim a distance 2d in still water, then relation between t1 , t2 & t3 .

Sol. Let v be the river velocity and u be the velocity of swimmer in still water. Then

$\large t_1 =  2 (\frac{d}{\sqrt{u^2 – v^2}}) $  …(i)

$\large t_2 = \frac{d}{u+v} + \frac{d}{u-v} = \frac{2ud}{u^2 -v^2}$ …(ii)

and , $\large t_3 = \frac{2d}{u}$  …(iii)

From equation (i), (ii) and (iii)

$\large t_1^2 = t_2  t_3$

$\large t_1 = \sqrt{t_2 t_3}$