2.8 Vertical motion under gravity  (Page 5/5)

The velocity of the ball at the highest point is zero. The only force on the ball is due to gravity. The accleration of ball all through out its motion is acceleration due to gravity “g”, which is directed downwards. The acceleration of the ball is constant and is not dependent on the state of motion - whether it is moving or is stationary.

The average velocity is ratio of displacement and time. Here, displacement is given. We need to find the time of travel. For the motion of ball, we consider the point of release as origin and upward direction as positive.

$$y=ut+\frac{1}{2}a{t}^2$$ $$\Rightarrow -45=0Xt+\frac{1}{2}X-10X{t}^2$$ $$\Rightarrow t^2=\frac{45}{5}=9$$ $$\Rightarrow t=\pm 3s$$

Neglecting negative time, t = 3 s. Magnitude of average velocity is :

$$v_{\text{avg}}=\frac{45}{3}=15m/s$$

On separation, ball acquires the velocity of elevator – not its acceleration. Once it is released, the only force acting on it is that due to gravity. Hence, acceleration of the ball is same as that due to gravity.

$$a=10\frac{m}{s^2}\text{downward}$$

On separation, ball acquires the velocity of elevator – not its acceleration. Once it is released, the only force acting on it is that due to gravity. Hence, acceleration of the ball is same as that due to gravity. The relative acceleration of the ball (considering downward direction as positive) :

$$a_{\text{rel}}=a_{\text{ball}}-a_{\text{elevator}}$$ $$\Rightarrow a_{\text{rel}}=10-\left(-5\right)=15\frac{m}{s^2}$$

The velocity of balloon is constant and is a measured value. Let the ball moves up with a velocity u. At the time of release, the pebble acquires velocity of balloon. For the motion of pebble, we consider the point of release as origin and upward direction as positive. Here,

$$y=vt=-uX5=-5u;u=u;a=-g;t=5s$$

Using equation for displacement :

$$y=ut+\frac{1}{2}a{t}^2$$ $$\Rightarrow -5u=uX5+\frac{1}{2}X-gX{5}^2$$ $$\Rightarrow 10u=5X25u=5\frac{m}{s}$$

The velocity of balloon is constant. Let the ball moves up with a velocity u. At the time of release, the pebble acquires velocity of balloon. For the motion of pebble, we consider the point of release as origin and upward direction as positive. Here,

$$y=?;u=10m/s;a=-g;t=5s$$

Using equation for displacement :

$$y=ut+\frac{1}{2}a{t}^2$$ $$\Rightarrow y=10X5+\frac{1}{2}X-gX{5}^2$$ $$\Rightarrow y=50-5X25$$ $$\Rightarrow y=-75m$$

Height of ballon when pebble is released from it,

$$\Rightarrow H=75m$$

We compare motion of two balls under gravity, when second ball is dropped. At that moment, two balls are 10 m apart. The first ball moves with certain velocity, whereas first ball starts with zero velocity.

Let us consider downward direction as positive. The velocity of the first ball when it reaches 10 m below the top is :

$$v^2=u^2+2ax$$ $$\Rightarrow v^2=0+2X10X5$$ $$\Rightarrow v=10m/s$$

Let the balls take time “t” to reach the gorund. First ball travels 10 m more than second ball. Let 1 and 2 denote first and second ball, then,

$$\Rightarrow 10+10t+\frac{1}{2}X10t^2=\frac{1}{2}X10t^2$$ $$10t=10$$ $$t=1s$$

In this time, second ball travels a distance given by :

$$\Rightarrow y=\frac{1}{2}X10t^2=5X12=5m$$

But, second ball is 15 m below the top. Hence, height of the top is 15 + 5 = 20 m.

For the motion of first ball dropped from the top, let downward direction be positive :

$$v_1=u+at$$ $$\Rightarrow v_1=at=10t$$

For the ball dropped from the top,

$$x=ut+\frac{1}{2}a{t}^2$$ $$\Rightarrow -20=0Xt+\frac{1}{2}X-10X{t}^2$$ $$\Rightarrow 20=5t^2$$ $$\Rightarrow t=\pm 2s$$

Neglecting negative value, t = 2s. Hence, velocity of the ball dropped from the top is :

$$v_1=10t=10X2=30m/s$$

For the motion of second ball projected from the bottom, let upward direction be positive :

$$v_2=u+\mathrm{at}$$ $$\Rightarrow v_2=u-10X2=u-20$$

Clearly, we need to know u. For upward motion,

$$x=ut+\frac{1}{2}a{t}^2$$ $$\Rightarrow 40=uX2+\frac{1}{2}X-10X{2}^2$$ $$\Rightarrow 40=2u-20=30m/s$$ $$\Rightarrow v_2=30-20=10m/s$$

Thus,

$$\Rightarrow \frac{v_1}{v_2}=\frac{20}{10}=\frac{2}{1}$$