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# A body is projected up with a velocity 96m/s. The ratio of the distance covered by the body in the last two seconds of its motion and last two seconds of its ascent.1) 1:72) 7:93)9:14)3:5

Last updated date: 13th Jun 2024
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Hint:- Imagine throwing a ball in the air. At some point of time when the ball reaches at the maximum height it can get then it will start to come down. One might have noticed that when the ball reaches at the top the K.E is converted into P.E and when it falls down P.E is converted into K.E. Use this relation to solve the question and also apply the equation of kinematics.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$;
S = Distance,
t = time;
a = acceleration;
u = initial velocity,
$mgh = \dfrac{1}{2}m{v^2}$;
m = mass;
g = Gravitational acceleration;
h = height;
$v = \dfrac{d}{t}$;
v = velocity;
d = displacement;
t = time;

Complete step-by-step solution:-

Step 1:
Calculate the distance covered in the last two seconds of its ascent.
The velocity in the last two seconds of its ascent would be 96m/s (given), as the body is in motion the whole time with the same velocity. We have been given time 2seconds. So, we can find the distance covered in 2 seconds by applying the equation of kinematics.
$s = ut + \dfrac{1}{2}a{t^2}$
Put in the given values:
$s = 96 \times 2 + \dfrac{1}{2} \times ( - 10) \times 4$ ….(Here g=-10, because the body is going against the gravity)
$s = 192 - 20$;
S = 172m;

Step 2: Calculate the distance covered in the last two seconds of the body motion.
Here when the body is at the maximum height, the body possesses potential energy (mgh). When the body just starts to fall down all the potential energy is converted to kinetic energy.
$mgh = \dfrac{1}{2}m{v^2}$;
The body has the same mass so it will cancel out.
$gh = \dfrac{1}{2}{v^2}$;
$2gh = {v^2}$;
Now put the value in the above equation:
$h = \dfrac{{{v^2}}}{{2g}}$;
Put the value;
$h = \dfrac{{{v^2}}}{{2 \times 10}}$;
$h = \dfrac{{96 \times 96}}{{20}}$;
The total distance is:
$h \approx 461$;
Now we know the total distance covered on going with a velocity of 96m/s. We can take out time.
$v = \dfrac{d}{t}$;
$t = \dfrac{d}{v}$;
Put the given values;
$t = \dfrac{{461}}{{96}}$;
Solve,
$t = 4.8s$;
Now we know the total time taken, velocity and total distance covered by the ball when falling from the maximum height.
Find out the distance covered by the ball in the last two seconds of its motion.
In 4.8 sec the body covers 461m;
So, in 2 seconds the distance the body covers will be = $\dfrac{{461 \times 2}}{{4.8}}$
D =$\dfrac{{461 \times 2 \times 10}}{{48}}$;
Calculate;
$D \approx 19m$;

Step 3: Find the ratio of the distances.
The ratio is$= \dfrac{S}{D}$;
Put the value in the above equation;
The ratio is$= \dfrac{{172}}{{19}}$;
Solve,
The ratio is$\approx \dfrac{9}{1}$;
The ratio is$\approx 9:1$;
Final Answer: Option”3” is correct. The ratio of the distance covered by the body in the last two seconds of its motion and last two seconds of its ascent is 9:1.

Note:- Here make sure to go step by step, be careful when making calculations, also make a relation between the distance covered and the velocity of the object by equating potential energy and kinetic energy together and then find out the distance covered in last two seconds of the motion of the body.