A ball is thrown up with velocity 19.6 m/s. The maximum height attained by the ball is

29.2 m
9.8 m
19.6 m
15.8 m

Solution

Let the maximum height attained by the ball be h.At maximum height ,velocity of ball, v = 0 Given, initial velocity, u= 19.6 m/s Using the equation of motion,

v²= u²+ 2gh

We get 0 =(19.6)²+ 2 (– 9.8) × h

⇒ h =\(\frac{(19.6)^{2}}{2\times 9.8}\)

= 19.6 m

Which one of the following equation represents the motion of a body moving with constant finite acceleration? In these equation, y denotes the displacement in time t and p, q and r are constant:

y= (p + qt )(t + pt)
y= p + t/r
y= (p + t) (q + t )(r +t)
\(y=\frac{(p+qt)}{rt}\)

Solution

Motion with constant acceleration is represented by a quadratic equation of t

Y = (p + qt) (r + pt) = pr + qrt + p²t + pqt²

The motion of particle is described by the equation x = a +bt², where a = 15 cm and b = 3 cm/sec². Its instant velocity at time 3 sec will be

36 cm/sec
9 cm/sec
4.5 cm/sec
18 cm/sec

Solution

A ball is released from the top of a tower of height h meters.It takes T seconds to reach the ground. What is the position of the ball at ^T⁄₃ second

\(\frac{8h}{3}\) meters from the ground
\(\frac{7h}{3}\) meters from the ground
^h⁄₉ meters from the ground
\(\frac{17h}{8}\) meters from the ground

Solution

A stone falls freely under gravity.It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is

\(h_{1}=\frac{h_{2}}{3}=\frac{h_{3}}{5}\)
h₂= 3h₁ and h₃ = 3h₂
h₁= h₂= h₃
h₁= 2h₂= 3h₃

Solution

A particle has initial velocity \(\left ( 2\overrightarrow{i}+3\overrightarrow{j} \right )\) and acceleration \(\left ( 0.3\overrightarrow{i}+0.2\overrightarrow{j} \right )\). The magnitude of velocity after 10 seconds will be

9√2 units
5√2 units
5 units
9 units

Solution

A particle moves a distance x in time t according to equationx= (t + 5)^–1. The acceleration of particle is proportional to

(velocity)^3/2
(distance)²
(distance)^-2
(velocity)^2/3

Solution

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18s. What is the value of v?(take g = 10 m/s²)

75 m/s
55 m/s
40 m/s
60 m/s

Solution

A man throws balls with same speed vertically upwards one after the other at an interval of 2 sec. What should be the speed of throw so that more than two balls are in air at anytime

only with speed 19.6 m/s
more than 19.6 m/s
at least 9.8 m/s
any speed less then 19.6 m/s.

Solution

Height attained by balls in 2 sec is

=¹⁄₂ × 9.8 × 4 =9.6m

the same distance will be covered in 2 second (for descent)Time interval of throwing balls, remaining same. So, for two balls remaining in air, the time of ascent or descent must be greater than 2 seconds. Hence speed of balls must be greater than 19.6m/sec.

A car moving with a speed of 40 km/hour can be stopped by applying brakes after at least 2m. If the same car is moving with a speed of 80km/hour, what is the minimum stopping distance.

Solution

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