Question

A particle is projected in upward direction with an initial velocity of $80ft/\sec$ . After how much time, it will be at a height of $96ft$ from the earth?
(A) $2{\text{ }}and{\text{ }}3{\text{ }}sec$
(B) $only{\text{ }}3{\text{ }}sec$
(C) $only{\text{ }}2{\text{ }}sec$
(D) $1{\text{ }}and{\text{ }}2{\text{ }}sec$

Answer 1

Hint : We know that when a particle is projected in the upward direction, it experiences the acceleration due to gravity. To solve this question, we will use the second equation of motion which gives the value of the height of the particle at a particular time. Thus, we will get a quadratic equation by solving which we will get the time taken by the particle to reach at a given height.
$h = {v_0}t - \dfrac{1}{2}g{t^2}$ , where, $h$ is the height, ${v_0}$ is the initial velocity, $g$ is the gravitational acceleration and $t$ is time.

Complete Step By Step Answer:
It is given that the particle is projected in upward direction with an initial velocity of $80ft/\sec$ , therefore, its initial velocity ${v_0} = 80ft/\sec$ .
We know that acceleration due to gravity $g = 32ft/{\sec ^2}$ .
We need to find the time when the particle is at the height $h = 96ft$ .
Now, we will apply the second equation of motion.
$h = {v_0}t - \dfrac{1}{2}g{t^2} \\\ \Rightarrow 96 = 80t - \dfrac{1}{2} \times 32{t^2} \\\ \Rightarrow 96 = 80t - 16{t^2} \\\ \Rightarrow 16{t^2} - 80t + 96 = 0 \\\ \Rightarrow {t^2} - 5t + 6 = 0 \\\ \Rightarrow {t^2} - 2t - 3t + 6 = 0 \\\ \Rightarrow t\left( {t - 2} \right) - 3\left( {t - 2} \right) = 0 \\\ \Rightarrow \left( {t - 2} \right)\left( {t - 3} \right) = 0 \\\$
$$ $\Rightarrow t - 2 = 0 \\\ \Rightarrow t = 2\sec \\\$ and $\Rightarrow t - 3 = 0 \\\ \Rightarrow t = 3\sec \\\$
Thus, the particle will be at a height of $96ft$ from the earth at $2{\text{ }}and{\text{ }}3{\text{ }}sec$ .
Hence, option A is the right answer.

Note :
Here, we are getting two values of time at which the particle is at the given height. This is because when we throw the particle in the upward direction, first the particle reaches at a given height during its upward motion. After that, it reaches its maximum height and then again it will be at the given height during its downward direction.