Question

From the top of a 490m high cliff, a boy throws a stone horizontally with an initial speed of 15m/s. What is the time taken by the ball to reach the ground?

Answer 1

In the question it is given to us that the boy throws the stone horizontally. Therefore we can imply that there is no initial vertically downward component of velocity given to the stone. It is also given to us that the boy throws the stone horizontally with an initial speed of 15m/s. hence we will use Newton’s kinematic equation to determine the time of flight of the stone.
Formula used:
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
Let us say a body is under uniform acceleration ‘a’. If the initial speed of the body is U, than at time ‘t’ the distance (S)covered by the body is given by Newton’s second kinematic equation as,
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$
In the above case the stone is under acceleration due to gravity i.e. g. hence the above equation becomes,
$S=Ut+\dfrac{1}{2}g{{t}^{2}}$
Let us first consider the vertical motion of the stone only. The stone is basically thrown from a height of 490m from the cliff. The initial velocity along the vertical of the stone is zero. Taking g as 10 meters per second square we get the time of flight of the stone in air i.e. time taken by the stone to reach the ground is equal to,
$\begin{aligned} & S=Ut+\dfrac{1}{2}g{{t}^{2}}\text{, }U=0 \\\ & \Rightarrow 490=\dfrac{1}{2}10{{t}^{2}} \\\ & \Rightarrow {{t}^{2}}=\dfrac{490}{5}=98{{\sec }^{2}} \\\ & \Rightarrow t=\sqrt{98{{\sec }^{2}}}=9.89\sec \\\ \end{aligned}$
Therefore the time taken by the ball to reach the ground is equal to 9.89 secs.

Note:
It is to be noted that the above time obtained is approximate as we have not taken into account the air drag. If we observe carefully the above solution, we can imply that the time of flight is independent of the initial velocity along the horizontal. To understand this consider a ball which is dropped from a height and from the same height we drop a ball but having some initial velocity along the horizontal. Both the balls will reach the ground at the same time. It is also to be noted that the horizontal velocity must be not that massive that it just never reaches the ground but keeps on rotating about the Earth.