Question

A toy rocket is launched with an initial velocity of 12.0m/s in the horizontal direction from the roof of a 37.0m tall building. The rocket's engine produces a horizontal acceleration of $2.2m/{s^2}$, in the same direction as the initial velocity, but in the vertical direction the acceleration is $9.8m/{s^2}$, downward. Air resistance can be neglected. How far does the rocket travel in a horizontal direction before it hits the ground? (in m).
A.40.4
B.40.9
C.40.7
D.40.3

Answer 1

To solve this question, we need to know the basic theory related to the motion of objects. As we know Horizontal distance covered by a rocket before hitting the ground is determined using the second equation of motion as discussed below. Here first we will calculate the Time for which rocket will be in air and then after calculate horizontal direction by using $s = ut + \dfrac{1}{2}a{t^2}$.

Complete answer:
Time for which rocket will be in air=$\sqrt {\dfrac{{2h}}{g}} = \sqrt {\dfrac{{2 \times 37}}{{9.8}}} = 2.7s$
Horizontal distance covered before hitting the ground. We have to apply second equation to solve this question
$\Rightarrow$ $s = ut + \dfrac{1}{2}a{t^2}$
Where S denotes displacement, u denotes initial velocity. a denotes acceleration and t denotes time.
Given,
initial velocity =12.0m/s
Time =2.7s
acceleration =$2.2m/{s^2}$
$12 \times 2.7 + 0.5 \times 2.2 \times {\left( {2.7} \right)^2} = 40.4m$
Thus, $40.4m$,rockets travel in a horizontal direction before they hit the ground.

So, the correct answer is “Option A”.

Note:
Always remember that the relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.