A man standing on the top of a tower 60m height throws a ball in the verticaly upward direction with a velocity 20m/s.the total distance travelled by the ball in moving upwards and returning to the thrower is
1. 10m
2. 20m
3. 30m
4. 40m
answer is 40m?

Question

Answer

At Highest point velocity will be zero.therefore from v²= u²+2gh0²= (20)² +2(-10)hon solving we will get h=20m here ‘h' is the distance travelled by ball in moving upwards.similarly the ball will travel 20 m during downward motion.so the answer is 20+20= 40m
Hope this will help.Thank You .

Answer

Ball should have initial velocity =0 and v=20m/s by applying v=u+at considering acceleration to be 10 time came out to be 2s . then put its value in s=ut+1/2atsquare swill came out to be 20m .. this is the distance of walf way till it goes up n as they ask for total distance it should double if it that is 40m ...

Answer

Understanding the Problem:
The problem involves a man standing on a tower who throws a ball vertically upward with a velocity of 20m/s. We need to calculate the total distance traveled by the ball in moving upwards and returning to the thrower.

Key Concepts:
- The ball's initial velocity is 20m/s upwards.
- The acceleration due to gravity is -9.8m/s^2.
- The total distance traveled by the ball will be the sum of the distances covered during the upward journey and the downward journey.

Calculating the Distance Traveled Upwards:
Using the kinematic equation:
v^2 = u^2 + 2as,
where v = final velocity (0 m/s at the highest point),
u = initial velocity (20 m/s),
a = acceleration (-9.8 m/s^2),
s = distance traveled upwards.

Plugging in the values, we get:
0 = (20)^2 + 2*(-9.8)*s,
0 = 400 - 19.6s,
19.6s = 400,
s = 400/19.6 ≈ 20.41m.

Calculating the Distance Traveled Downwards:
The distance traveled downwards will be the same as the distance traveled upwards. So, it is also approximately 20.41m.

Total Distance Traveled:
Adding the distances covered during the upward and downward journeys:
Total distance = 20.41m + 20.41m = 40.82m ≈ 40m.

Therefore, the total distance traveled by the ball in moving upwards and returning to the thrower is approximately 40m.

Similar JEE Doubts

A man standing on the top of the tower 60m height throws a ball in the verticaly upward direction with a a velocity of 20m/s.how long will it take for the ball to pass by the man while moving in the downward direction? 1. 2 sec 2. 4 sec 3. 6 sec 4. 8 sec? given answer is 4 sec?

A boy standing on a long railroad car throws a ball straight upward. The car is moving on the horizontal road with an acceleration of 1m/s2 and the projection velocity in the vertical direction is 9.8m/s. How far behind the boy will the ball fall on the car?

An electron starting from rest has a velocity that increases linearly with the time that is v equal to kt,where k is 2m/sec2.the distance travelled in the first 3 sec will be 1)9m 2)16m 3)27m 4)36m?

two projectiles are thrown from the same point simultaneously with same velocity 10m/s .one goes straight vertically while other at 60 degree with the vertical. what will be the distance of separation between the two after 1 second of their throw?

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Chemistry for JEE Main & Advanced

Mathematics (Maths) for JEE Main & Advanced

Chapter-wise Tests for JEE Main & Advanced

Top Courses for JEE

Top Courses for JEE

Suggested Free Tests

Related Content

JEE Courses

Mock Test Series

Past Year Papers

Additional Courses

Company