A plane flying horizontally at a height of 1500 m with a velocity of 200 ms^-1 passes directly overhead on antiaircraft gun. Then the angle with the horizontal at which the gun should be fired from the shell with a muzzle velocity of 400 ms^-1 to hit the plane, is

90°
60°
30°
45

Solution

Horizontal distance covered should be same for the time of collision.

400 cos θ=200 or cos θ=¹⁄₁ or θ=60°

A body is thrown horizontally with a velocity \(\sqrt{2gh}\) from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower.The value of x is

gh
\(\frac{gh}{2}\)
2h
\(\frac{2gh}{3}\)

Solution

The equation of a projectile is

\(y=\sqrt{3}x-\frac{gx^{2}}{2}\)

The angle of projection is given by

tan θ=¹⁄_√3
tan θ =√3
\(\frac{\pi }{2}\)
zero.

Solution

A cricket ball is hit with a velocity 25 ms^-1, 60° above the horizontal. How far above the ground, ball passes over a fielder 50 m from the bat (consider the ball is struck very close to the ground)?

Take √3=1.7 and g= 10 ms^-2

6.8 m
7 m
5 m
10 m

Solution

Two bullets are fired horizontally, simultaneously and with different velocities from the same place. Which bullet will hit the ground earlier ?

It would depend upon the weights of the bullets.
The slower one.
The faster one.
Both will reach simultaneously.

Solution

The initial velocity in the vertically downward direction is zero and same height has to be covered.

An aeroplane flying at a constant speed releases a bomb.As the bomb moves away from the aeroplane, it will

always be vertically below the aeroplane only if the aeroplane was flying horizontally
always be vertically below the aeroplane only if the aeroplane was flying at an angle of 45° to the horizontal
always be vertically below the aeroplane
gradually fall behind the aeroplane if the aeroplane was flying horizontally.

Solution

Since horizontal component of the velocity of the bomb will be the same as the velocity of the aeroplane,therefore horizontal displacements remain the same at any instant of time.

Three particles A, B and C are thrown from the top of a tower with the same speed.A is thrown up,B is thrown down and C is horizontally. They hit the ground with speeds v_A, v_B and v_C respectively then,

v_A= v_B= v_C
v_A= v_B > v_C
v_B > v_C > v_A
v_A > v_B = v_C

Solution

For A: It goes up with velocity u will it reaches its maximum height (i.e. velocity becomes zero) and comes back to O and attains velocity u.

A person aims a gun at a bird from a point at a horizontal distance of 100 m. If the gun can impact a speed of 500 ms^-1 to the bullet. At what height above the bird must he aim his gun in order to hit it? (g= 10 ms^-2)

10.4 cm
20.35 cm
50 cm
100 cms

Solution

A projectile thrown with velocity v making angle θ with vertical gains maximum height H in the time for which the projectile remains in air, the time period is

\(\sqrt{Hcos\theta /g}\)
\(\sqrt{Hcos\theta /g}\)
\(\sqrt{4H/g}\)
\(\sqrt{4H/g}\)

Solution

A large number of bullets are fired in all directions with the same speed v. What is the maximum area on the ground on which these bullets will spread?

\(\frac{\pi v^{2}}{g}\)
\(\frac{\pi v^{4}}{g^{2}}\)
\(\pi^{2}\frac{ v^{2}}{g^{2}}\)
\(\frac{\pi^{2} v^{4}}{g^{2}}\)

Solution

Maximum possible horizontal range= v²/g

Maximum possible area of the circle

\(=\pi \left ( \frac{V^{2}}{g} \right )^{2}=\frac{\pi v^{4}}{g^{2}}\left [ Here\, r=\frac{V^{2}}{g} \right ]\)

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