Question

a person falling down freely shoots a ball upward with a force of 100N. The mass of ball is 100grams. If the person reaches the ground in 5 sec and the person shoots the ball 2 sec after jumping find the time taken by ball to reach the ground.​

Answer 1

To solve this problem, we can break it down into two parts: the motion of the person falling and the motion of the ball being shot upward.

1. The motion of the person falling: Given that the person reaches the ground in 5 seconds, we know that the time of flight for the person falling is 5 seconds.
2. The motion of the ball being shot upward: The ball is shot 2 seconds after the person jumps. This means the ball has 5 - 2 = 3 seconds of flight time.

To find the time taken by the ball to reach the ground, we need to consider the total time of flight for the ball. Since the ball is shot upward, it will reach its highest point and then fall back to the ground.

The time taken for the ball to reach its highest point will be half of its total flight time. Therefore, the ball reaches the highest point is 3 / 2 = 1.5 seconds.

The time taken for the ball to fall back down to the ground is also 1.5 seconds, as the time of ascent is equal to the time of descent.

Thus, the total time taken by the ball to reach the ground is 1.5 seconds (time to reach the highest point) + 1.5 seconds (time to fall back down) = 3 seconds.

Therefore, the ball takes 3 seconds to reach the ground after being shot upwards by the person.

Answer 2

Suppose they meet after time t

0

​

then sum of magnitudes of their displacement should be H=98m

or h

1

​

+h

2

​

=(49t

0

​

−

2

9.8

​

t

0

2

​

)+

2

9.8

​

t

0

2

​

=98 so t

0

​

=2sec

Now we know that after 2sec the velocity of first particle in upward direction will be v

1

​

=49−9.8×2=29.4m/s

and the velocity of second particle after these 2sec will be v

2

​

=9.8×2=19.6m/s in downward direction.

so the net momentum in upward direction will be p=m×29.4−m×19.6=9.8m

so net upward velocity of htis combined mass(2m) will be V=

2m

p

​

=4.9m/s

and this height from ground will be h

1

​

=49×2−4.9×4=78.4m

now using following equation h=−Vt+4.9t

2

putting h=78.4m and V=4.9m/s

we get t=4.5sec

so time of flight will be t+t

0

​

=6.5sec

Solve any question of Motion in a Straight Line with:-

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