CameraIcon

SearchIcon

MyQuestionIcon

3

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Derivation of Velocity-Time Relation by Graphical Method

A ball is pro...

Question

A ball is projected vertically up such that it passes through a fixed point after a time $t_{1}$ and $t_{2}$ , respectively. Find
a.The height at which the point is located with respect to the point of projection
b. The speed of the projection of the ball.
c.The velocity of the ball at the time of passing through point P.
d. (i) The maximum height reached by the ball relative to the point of projection A and (ii) maximum height reached by the ball relative to point P under consideration.
e. The average speed and average velocity of the ball during the motion from A to P for the time $t_{1}$ and $t_{2}$ , respectively.

Open in App

Solution

a,b. Let the ball be projected up with an initial velocity u. It passes through point P at t = $t_{1}$ during its ascent and at t = $t^{2}$ during its descent.
For the motion of the ball from A to P,
s = +h, $v_{0}$ = u, a = -g, and t = $t_{1}$ and $t_{2}$
Substituting the above value, in s = $v_{0} t + (1 / 2) a t^{2}$ , we get
h = $u t_{1}$ - (1/2) $g t_{1}^{2}$ (upward motion)............(i)
= $u t_{2}$ - (1/2) $g t_{2}^{2}$ (downward motion)..........(ii)
Hence, from the above equations, $u t_{1}$ - (1/2) $g t_{1}^{2}$ = $u t_{2}$ - (1/2) $g t_{2}^{2}$
u( $t_{2} - t_{1}$ ) = $\frac{1}{2} g (t_{2} - t_{1}) (t_{2} + t_{1})$ $\Rightarrow$ $u = \frac{1}{2} g (t_{2} + t_{1})$
Substituting the value of u in (i) and (ii), we get h = (1/2)g $t_{1} t_{2}$
Solving the above equation,we have
h = $\frac{g t_{1} t_{2}}{2}$ and $v = \frac{g (t_{1} + t_{2})}{2}$
c. Using the relation $\to v = \to u + \to a t$
$v_{p} = (\frac{g (t_{2} + t_{1})}{2})$ - $g t_{1} = \frac{g}{2} (t_{2} - t_{1})$
d.(i). Maximum height reached by abll from the point of projection.
Using $v^{2} = u^{2} + 2 a s$
0 = ${(\frac{g}{2 (t_{2} + t_{1}})}^{2}$ - 2gH $\Rightarrow H_{m a x} = \frac{g}{8} (t_{1} + t_{2})^{2}$
(ii). Height reached by ball from point P,
$h^{'} = H_{m a x} - h = \frac{g}{8} (t_{1} + t_{2})^{2} - \frac{1}{2} g t_{1} t_{2}$ = $\frac{g}{8} (t_{2} - t_{1})^{2}$
e.(i). A to P for time $t_{1}$
The average speed and average velocity from A to P will be same as the particle is moving in straight line without changing the direction.
< v > = $\frac{Δ y}{Δ t} = \frac{h}{t_{1}} = \frac{(1 / 2) g t_{1} t_{2}}{t_{1}} = \frac{21}{2} t_{2}$
(ii) Motion from A to P for time $t_{2}$ .
Average speed = $\frac{T o t a l d i s t a n c e}{T o t a l t i m e}$
= $\frac{h + 2 h^{'}}{t_{2}}$
= $\frac{(1 / 2) g t_{1} t_{2} + 2 (\frac{g}{8} (t_{1} + t_{2})^{2})}{t_{2}}$
= $\frac{g (t_{1}^{2} + t_{2}^{2})}{4 t_{2}}$
Motion from A to P for time $t_{2}$
Average velocity < $\to v$ > = $\frac{Δ \to y}{Δ \to t} = \frac{h}{t_{2}} = \frac{(1 / 2) g t_{1} t_{2}}{t_{2}} = \frac{1}{2} g t_{1}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A cricket ball thrown across a field is at heights

h_{1}

h_{2}

from the point of projection at times

t_{1}

t_{2}

respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is:

Q. A ball is thrown vertically upward from the top of a tower. Velocity at depth

h

from the point of projection is twice of the velocity at height

h

above the point of projection. Find the maximum height reached by the ball.

Q. A ball is thrown vertically upwards from the top of a tower. Velocity at a point

^{'} h^{'} m

vertically below the point of projection is twice the downward velocity at a point

^{'} h^{'} m

vertically above the point of projection. The maximum height reached by the ball above the top of the tower is:

Q. A ball projected from ground vertically upward is at same height at time

t_{1}

t_{2}

. The speed of projection of ball is: [Neglect the effect of air resistance].

Q. A body is projected up with velocity

u

. It reaches the same point in its path at

t_{1}

t_{2}

from the time of projection. Then

t_{1} + t_{2}

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equations of Motion

PHYSICS

Watch in App

explore_more_icon

Explore more

Derivation of Velocity-Time Relation by Graphical Method

Standard IX Physics

Join BYJU'S Learning Program

CrossIcon