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A person, standing on the roof of a 40 m...

A person, standing on the roof of a 40 m high tower, throws a ball vertically upwards with speed 10 m/s. Two seconds later, he throws another ball again in vertical direction (use `g=10(m)/(s^2)`). Both the balls hit the ground simultaneously.

A

The first ball hits the ground after 4 seconds.

B

The second ball was projected vertically downwards with speed 5 m/s

C

The second ball was projected vertically downwards with speed 5 m/s.

D

the distance travelled by the first ball is 10 m greater than the distance travelled by the second ball.

Text Solution

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To solve the problem, we need to analyze the motion of both balls thrown from the top of a 40 m high tower. The first ball is thrown upwards with an initial speed of 10 m/s, and the second ball is thrown 2 seconds later. Both balls hit the ground simultaneously. ### Step 1: Calculate the time taken by the first ball to hit the ground. The first ball is thrown upwards with an initial speed \( u_1 = 10 \, \text{m/s} \). The height of the tower is \( h = 40 \, \text{m} \). We will use the equation of motion: \[ S = ut + \frac{1}{2} a t^2 \] Here, \( S \) is the displacement (which will be -40 m since it moves downward), \( u \) is the initial velocity, \( a \) is the acceleration (which is -10 m/s² due to gravity), and \( t \) is the time taken to hit the ground. Substituting the values: \[ -40 = 10t - \frac{1}{2} \cdot 10 \cdot t^2 \] This simplifies to: \[ -40 = 10t - 5t^2 \] Rearranging gives: \[ 5t^2 - 10t - 40 = 0 \] Dividing the entire equation by 5: \[ t^2 - 2t - 8 = 0 \] ### Step 2: Solve the quadratic equation. Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1, b = -2, c = -8 \): \[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \] \[ t = \frac{2 \pm \sqrt{4 + 32}}{2} \] \[ t = \frac{2 \pm \sqrt{36}}{2} \] \[ t = \frac{2 \pm 6}{2} \] This gives us two solutions: \[ t = \frac{8}{2} = 4 \, \text{s} \quad \text{and} \quad t = \frac{-4}{2} = -2 \, \text{s} \] Since time cannot be negative, we have: \[ t_1 = 4 \, \text{s} \] ### Step 3: Calculate the time taken by the second ball to hit the ground. The second ball is thrown 2 seconds after the first ball, so it will hit the ground after: \[ t_2 = t_1 - 2 = 4 - 2 = 2 \, \text{s} \] ### Step 4: Determine the initial velocity of the second ball. Since both balls hit the ground simultaneously, we can use the same equation of motion for the second ball. The second ball is thrown downwards with an unknown initial velocity \( u_2 \): Using the same displacement equation: \[ -40 = u_2 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) \] This simplifies to: \[ -40 = 2u_2 - 20 \] Rearranging gives: \[ 2u_2 = -40 + 20 \] \[ 2u_2 = -20 \] \[ u_2 = -10 \, \text{m/s} \] This means the second ball is thrown downwards with a speed of 10 m/s. ### Summary of Results - The first ball takes 4 seconds to hit the ground. - The second ball takes 2 seconds to hit the ground and is thrown downwards with a speed of 10 m/s.
To solve the problem, we need to analyze the motion of both balls thrown from the top of a 40 m high tower. The first ball is thrown upwards with an initial speed of 10 m/s, and the second ball is thrown 2 seconds later. Both balls hit the ground simultaneously. ### Step 1: Calculate the time taken by the first ball to hit the ground. The first ball is thrown upwards with an initial speed \( u_1 = 10 \, \text{m/s} \). The height of the tower is \( h = 40 \, \text{m} \). We will use the equation of motion: \[ S = ut + \frac{1}{2} a t^2 ...
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