Answer
Verified
91.5k+ views
Hint The second equation of motion can be used to solve the given question. The height of the building is constant. Thus equating the distances covered by both balls can give the time at which they meet.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the distance covered by the body.
$u$ is the initial velocity of the body.
$v$ is the final velocity of the body.
$a$ is the acceleration.
$t$ is the time taken.
Complete Step by step solution
It is given in the question that,
The height of the building, $h = 80m$
Ball A is dropped from the top of the building with initial velocity of, ${u_A} = 0m/s$
Ball B is thrown upwards from the bottom of the building with initial velocity of, ${u_B} = 50m/s$
We know that the only force acting here is gravity therefore the acceleration of both balls is $g = 9.8m/s$ in the downward direction.
When the balls meet, the sum of the distances traveled by them totals to the height of the building.
Let, distance traveled by ball A be, ${s_A}$.
And the distance traveled by ball B be, ${s_B}$.
Then, $h = {s_A} + {s_B}$
From the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
For ball A,
$u = 0$
$a = 9.8m{s^{ - 2}}$
Therefore,
${s_A} = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_A} = \dfrac{1}{2} \times 9.8{t^2}$
For ball B,
$u = 50m$
$a = - 9.8m/{s^2}$ (it is negative because the gravity tries to retard the motion of the ball)
Therefore,
${s_B} = 50 \times t - \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_B} = 50t - \dfrac{1}{2} \times 9.8 \times {t^2}$
Adding the distances covered by both balls,
$h = {s_A} + {s_B}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = 50t$
$t = \dfrac{{80}}{{50}} = 1.6$
The time when both balls meet is at$1.6\sec $.
Therefore option (1) is correct.
Note Velocity and acceleration are vector quantities. They can be positive or negative, depending on their directions. Velocity of an object may be negative when the object moves in the negative direction. But acceleration can be negative in two cases, first when the acceleration is in the opposite direction, or when the object is retarding.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the distance covered by the body.
$u$ is the initial velocity of the body.
$v$ is the final velocity of the body.
$a$ is the acceleration.
$t$ is the time taken.
Complete Step by step solution
It is given in the question that,
The height of the building, $h = 80m$
Ball A is dropped from the top of the building with initial velocity of, ${u_A} = 0m/s$
Ball B is thrown upwards from the bottom of the building with initial velocity of, ${u_B} = 50m/s$
We know that the only force acting here is gravity therefore the acceleration of both balls is $g = 9.8m/s$ in the downward direction.
When the balls meet, the sum of the distances traveled by them totals to the height of the building.
Let, distance traveled by ball A be, ${s_A}$.
And the distance traveled by ball B be, ${s_B}$.
Then, $h = {s_A} + {s_B}$
From the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
For ball A,
$u = 0$
$a = 9.8m{s^{ - 2}}$
Therefore,
${s_A} = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_A} = \dfrac{1}{2} \times 9.8{t^2}$
For ball B,
$u = 50m$
$a = - 9.8m/{s^2}$ (it is negative because the gravity tries to retard the motion of the ball)
Therefore,
${s_B} = 50 \times t - \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_B} = 50t - \dfrac{1}{2} \times 9.8 \times {t^2}$
Adding the distances covered by both balls,
$h = {s_A} + {s_B}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = 50t$
$t = \dfrac{{80}}{{50}} = 1.6$
The time when both balls meet is at$1.6\sec $.
Therefore option (1) is correct.
Note Velocity and acceleration are vector quantities. They can be positive or negative, depending on their directions. Velocity of an object may be negative when the object moves in the negative direction. But acceleration can be negative in two cases, first when the acceleration is in the opposite direction, or when the object is retarding.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main