Test: Uniformly Accelerated Motion - JEE MCQ

v² = u² + 2as

Where:
v = final velocity of the object
u = initial velocity of the object
a = acceleration of the object (in this case, due to gravity)
s = displacement of the object

In this scenario, the boy drops the stone from rest (u = 0) and it falls freely under the influence of gravity (a = 10 m/s²). The displacement (s) is given as the height of the tower, which is 20 m.

Let's calculate the final velocity (v) using the given values:

v² = u² + 2as

Since the stone is dropped from rest, u = 0:

v² = 0² + 2 * 10 * 20

v² = 0 + 400

v² = 400

Taking the square root of both sides:

v = √400

v = 20 m/s

Therefore, the velocity with which the stone hits the ground is 20 m/s.

So, the correct answer is option (A) 20 m/s.

To find the maximum height attained by a body thrown upwards with a velocity of 15 m/s, we can use the equation of motion for vertical motion:

v² = u² + 2as

where:
v = final velocity (which is 0 m/s at the highest point)
u = initial velocity (15 m/s)
a = acceleration due to gravity (10 m/s², assuming upward as positive)
s = displacement (height)

Substituting the values into the equation, we get:

0² = (15 m/s)² + 2 * (-10 m/s²) * s

Simplifying the equation:

0 = 225 m²/s² - 20 m/s² * s

20 m/s² * s = 225 m²/s²

s = 225 m²/s² / 20 m/s²

s = 11.25 m

Therefore, the maximum height attained by the body is 11.25 meters. The correct solution is option A.

According to equation of motion, distance covered in nth sec.
S_n = u = a/2 (2n - 1)
S_n = a/2 (2n - 1)
∴ S₁ : S₂ : S₃ = {2(1)−1} : {2(2)−1} : {2(3)−1}
= 1 : 3 : 5

At maximum height its velocity is zero and the acceleration on it, is the acceleration due to gravity which is a constant quantity.

Let the height of the tower is h.
The motion of the ball is constant accelerating motion with acceleration g = 10m/s².
Initial velocity of ball, u = 20 m/s
Final velocity of ball, v = 80 m/s
According to equation of motion for constant accelerating motion,
(Here, downward direction is negative and upward direction is positive)
v² − u² = 2(−g)(−h)
⇒ v² − u² = 2gh
⇒ (80)² − (20)² = 2(10)h
⇒ 6400 − 400 = 20h
⇒ 6000 = 20h
⇒ h = 300m

For a ball thrown vertically upward and returning to the initial position under the influence of gravity:
Time of ascent = Time of descent = 5 sec

At a car travelling at a speed of 30km/h is brought to rest in a distance of 8m by applying brakes if the same car is moving at a speed of 60km/h then that it can be brought to rest with same brakes in:
1/2 mv² = Fs
on the doubling, the speed and the required distance will become 4 times that is s = 4 × 8 = 32m

u = 10m/s, v = 20m/s, x = 130m

Case 1: Lift is stationery

Given: The initial speed of the ball is 49 m/s and the life is stationery.
The second equation of motion is given as,
s = ut + 1/2 at²

Where s is the displacement of the ball, u is the initial velocity, t is the time taken by the ball and a is the acceleration of the ball in the lift.
In this case, displacement is 0 since the ball comes back to the boy who is standing on the stationary lift.
By substituting the given values in the above equation, we get
0 = 49 × t − 1/2 × 9.8 × t²
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.

Case 2: The lift is moving with a uniform velocity of 5 m/s

The lift is moving with a uniform velocity of 5 m/s in the upward direction which means there is no acceleration in the lift.
So in this case also, the relative speed of the ball with respect to the boy remains the same that is 49 m/s.
Therefore, the time taken by the ball to return to the boy's hand will same as in case I.
Thus, the time taken by the ball to return to the boy's hands when the lift is stationary is 10 s.

Step 1: Equations of motion for Last second

Let the total time of fall is 't' s, in the last second (t^th second) displacement is 9h/25 Initial speed u=0 ;  Acceleration a = g
Using expression for displacement in n^th second (Taking downward positive)

Step 2: Equations of motion for whole motion
(Taking downward Positive)

Step 3: Equation solving
From equation (1) and (2)

From equation (2)

Hence the value of h is 125 m.