Question

A particle is thrown from top of a building of height $50m$ with a velocity $30m/s$ at an angle of $30^\circ$ above the horizontal then the distance from the base of building to the point where the particle hits the ground is.

Answer 1

We solve this question using the formula for range in a horizontal projectile. The only unknown from the horizontal projectile range formula is the time of flight. We first find the time of flight using the second equation of motion and then substitute it in the formula of range in horizontal projectile motion.

Formula used:
Second equation of motion is $s = ut + \dfrac{1}{2}a{t^2}$
Here, distance is represented by $s$
Initial velocity is represented by $u$
Time is represented by $t$
Acceleration is represented by $a$
Range in horizontal projectile is equal to $R = u\cos \theta \times t$
Here, Range is represented by $R$
Horizontal velocity is represented by $u$
Angle of projection made with the horizontal is represented by $\theta$

Complete step by step answer: We take sign conventions under consideration to solve this question. Any force acting upwards is positive and downwards is negative.
From the second equation of motion $s = ut + \dfrac{1}{2}a{t^2}$
The distance travelled by the particle vertically is nothing but the height of the building $s = 50m$
The initial velocity is taken as the vertical component of the velocity at which the particle is thrown $u = 30\sin 30^\circ = 15u$
We take the angle as $30^\circ$ because it is the angle at which the particle is thrown.
Gravity is the only acceleration acting on the particle, since acceleration due to gravity is acting downwards, we take it as a negative sign $a = - g = - 9.8m/s$
Substituting in the equation $- 50 = - 15t + \dfrac{1}{2}( - g){t^2}$
The particle is falling hence the height from which it is falling is taken negative.
Acceleration due to gravity is equal to $g = 9.8m/s$
The finial equation becomes $- 50 = 15t - 0.5 \times 9.8 \times {t^2}$
Solving for time we get $t = 5\sec ,t = - 2\sec$
Time cannot be negative so time of flight is taken as $t = 5\sec$
Substituting this in the formula of range in horizontal projectile $R = u\cos 30^\circ t$
$R = 30 \times \cos 30^\circ \times 5$
$R = 130m$

Hence the horizontal distance travelled by the particle is $R = 130m$

Note: We cannot use the direct formula of range in this question because the height at which the particles initially is and the final height of the particle are not the same. We have to find the time of flight and then use it in a range formula. We can also find the time of flight by adding time of accent plus time of descent. Students might forget to sign convention into consideration. It is a must to take signs under consideration.