Question

A particle is projected up with initial speed u = 10 $m{s}^{-1}$ from the top of a building at time t = 0 as shown in the figure. At time t = 5 s the particle strikes the ground. Find the height of the building.

Answer 1

Using the scalar method

First, the particle moves up and reaches the highest point. Then the particle moves towards downward direction and finally strikes ground after 5 s.

At the highest point, the speed of the particle will be zero.

Here speed is in upward direction and acceleration is in the downward direction. Hence, the sign of acceleration should be negative.

Using $v^2=u^2+2as$

Hence, 0 = $(10{)}^2-2\times 10\times H_1$ $\Rightarrow$ $H_1$ = 5m

Using v = u + at from A to B,

0 = 10 - 10 $\times t_1$ $\Rightarrow$

$t_1$ = 1s

At highest point, the velocity of the particle will (a) be zero.

The particle starts moving towards downward direction. Here the velocity and acceleration both are in downward direction. Hence, the sign of acceleration should be positive.

The time taken by the particle from B to C will be 5 - 1 = 4 s

Now using s =ut +$\frac{1}{2}a{t}^2$ from B to C

$(H_1+H)=\left(1/2\right)\times 10\times 4^2$ = 80m

5 + H = 80m

$\Rightarrow$ Hence height of the building H = 75m

Using vector method

The particle starts from A and finally reaches at C.

Let us take the origin at A. The upward direction is taken as positive and the downward direction is taken as negative.

The particle moves in a gravitational field where the acceleration due to gravity is always acting in the downward direction whether it is moving upward or downward.

Hence, acceleration vector $\overrightarrow{a}$ will always be -10 $ms-2$, as its magnitude, as well as direction, remain constant always throughout the motion.

Hence, acceleration $\overrightarrow{a}$ = -10 $m{s}^{-2}$

Initially at t = 0, the particle is projected in the upward direction. Hence, initial velocity $\overrightarrow{u}$ = 10 $ms-1$. The particle moves from A to B (upward) and then B to C (downward).

The motion of the particle from A to B then again passes point A. The net displacement of the particle upto this instant is zero. Then particle crosses point A and finally reaches to C. We know net displacement is equal to the difference of final position vector and the initial position vector. Hence, net displacement of the particle during motion (t = 5 s) is -H (m).

Using $\overrightarrow{s}=\overrightarrow{u}t+\left(1/2\right)\overrightarrow{a}{t}^2$

=$\left(10\right)\times 5+\left(1/2\right)(-10)(5{)}^2$ = 50 - 125 = - 75 (m)

Hence, H = 75m.