Question

From the top of a tower of height 20 m, two balls are projected, one with speed 20 m/s at an angle of 30° with horizontal in upward direction and second with speed 20 m/s at an angle of 30° with horizontal in downward direction. Distance between the balls, when they land on the ground is

Answer 1

Explanation:

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Answer 2

Given,

• A ball dropped from a tower in an upward direction with a speed of 20m/s making angle 30° with the horizontal.

• A ball dropped with speed 20m/s in the downward direction making an angle 30° with the horizontal.

To find,

Distance between the ball when they both lay on the ground.

Solution,

Considering the ball travelling upwards.

Horizontal velocity = vCosθ = $10\sqrt{3}m/s$

Vertical velocity = vSinθ = $10m/s$

To calculate the horizontal distance, we need to calculate the time taken to reach the ground.

$s = ut + \frac{1}{2}at^{2} \\-20 = 10t - \frac{1}{2}gt^{2}\\t = 1 + \sqrt{5} , 1 - \sqrt{5}$

We will consider, t = $1 + \sqrt{5}$

Horizontal distance travelled by ball 1 = $10\sqrt{3}$$( 1 + \sqrt{5} )m$

Considering the ball travelling downwards.

Horizontal velocity = vCosθ = $10\sqrt{3}m/s$

Vertical velocity = vSinθ = $10m/s$

To calculate the horizontal distance, we need to calculate the time taken to reach the ground.

$s = ut + \frac{1}{2}at^{2} \\20 = 10t + \frac{1}{2}gt^{2}\\t = -1 + \sqrt{5} , -1 - \sqrt{5}$

We will consider, t = $-1 + \sqrt{5}$

Horizontal distance travelled by ball 1 = $10\sqrt{3}$$( -1 + \sqrt{5} )m$.

Difference between the positions  = $10\sqrt{3}$$( 1 + \sqrt{5} )m$ - $10\sqrt{3}$$( -1 + \sqrt{5} )m$

                                                          $= 20\sqrt{3}m$

Therefore, the distance between the two balls will be $20\sqrt{3}m$.