A missile is fired for maximum range with an initial velocity of 20 m/s. If g =10 m/s², the range of the missile is

40 m
50 m
60 m
20 m

Solution

For maximum range,the angle of projection, θ = 45°.
\(∴R=\frac{u^{2}sin2\theta }{g}=\frac{(20)^{2}sin(2\times 45^{\circ})}{10}=\frac{400\times 1}{10}=40m\)

If 0.5 i + 0.8 j + ck is a unit vector, then c is

\(\sqrt{(0.89)}\)
0.2
0.3
\(\sqrt{(0.11)}\)

Solution

\(\sqrt{(0.5)^{2}+(0.8)^{2}+c^{2}}=1\)

(0.89+c²)=(1)²=1

c²=1-0.89=0.11 or c=\(\sqrt{(0.11)}\)

A boat which has a speed of 5 km h^-1 in still water crosses a river of width 1km along the shortest possible path in 15 minutes. The velocity of the river water is

1 km h^-1
3 km h^-1
4 km h^-1
\(\sqrt{41}kmh^{-1}\)

Solution

\(v=\frac{1km}{\frac{1}{4}h}-4kmh^{-1},v_{b}=5kmh^{-1}\)

\(v_{w}=\sqrt{v_{b}^{2}-v^{2}}=\sqrt{25-16}=\sqrt{9}=3kmh^{-1}\)

The ratio of angular speeds of minutes hand and hour hand of a watch is

1 : 12
6 : 1
12 : 1
1 : 6

Solution

Angular speed of hour hand,\(\omega _{1}=\frac{2\pi }{(12\times 60)}\)

Angular speed of minute hand \(\omega _{2}=\frac{2\pi }{60}\)

\(∴\frac{\omega _{2}}{\omega _{1}}=\frac{12\times 60}{60}=\frac{12}{1}\)

In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity u, the horizontal range is

\(\frac{u^{2}}{g}\)
\(\frac{u^{2}}{g}\)
\(\frac{u^{2}}{3g}\)
\(\frac{u^{2}}{3g}\)

Solution

\(\theta =45^{\circ},R=\frac{u^{2}}{g}\)

At the highest point on the trajectory of a projectile,its

potential energy is minimum
kinetic energy is maximum
total energy is maximum
kinetic energy is minimum.

Solution

Velocity and kinetic energy is minimum at the highest point.

\(K.E=\frac{1}{2}mv^{2}\: cos^{2}\theta\)

The time of flight of a projectile on an upward inclined plane depends upon

angle of inclination of the plane
angle of projection
the value of acceleration due to gravity
all of the above.

Solution

\(T=\frac{2u\: sin(\theta -\alpha )}{g\: cos\: \alpha }\)

In uniform circular motion, the velocity vector and acceleration vector are

perpendicular to each other
same direction
opposite direction
not related to each other

Solution

In uniform circular motion speed is constant. So,no tangential acceleration.
It has only radial acceleration \(a_{R}=\frac{V^{2}}{R}\)
[directed towards center]
and its velocity is always in tangential direction. So thesetwo are perpendicular to each other

A body is thrown with a velocity of 9.8 ms^–1 making an angle of 30º with the horizontal. It will hit the ground after a time

3.0 s
2.0 s
1.5 s
1 s

Solution

Time of flight=\(\frac{2 \: usin\: \theta }{g}\)

\(=\frac{29\times 8\times sin30º}{9.8}=2\times \frac{1}{2}=1sec\)

If \(\dot{a}=\hat{i}-\hat{j}+\hat{k}\) and\(\dot{b}=2\hat{i}-\hat{j}+3\hat{k}\) then the unit vector along \(\underset{a}{\rightarrow}+\underset{b}{\rightarrow}\) is

\(\frac{3i+4k}{5}\)
\(\frac{3i+4k}{5}\)
\(\frac{-3i-4k}{5}\)
None of these

Solution

\(\dot{a}+\dot{b}=3\dot{i}+4k\)

∴Required unit vector

\(=\frac{\dot{a}+\dot{b}}{\left | \dot{a}+\dot{b} \right |}=\frac{3\dot{i}+4\hat{k}}{5}\)

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