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In previous question, the horizontal dis...

In previous question, the horizontal displacement (from the foot of the tower) of the ball is approximately equal to

A

50 m

B

60 m

C

70 m

D

80 m

Text Solution

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The correct Answer is:
To solve the problem, we need to find the horizontal displacement of a ball projected from the top of a 40 m high tower at an initial speed of 20 m/s at an angle of 30 degrees. We will break down the solution into steps. ### Step 1: Determine the time of flight (T1) The total time taken for the ball to hit the ground can be found using the vertical motion equations. The height of the tower is 40 m, and we can use the following kinematic equation: \[ S = ut + \frac{1}{2} a t^2 \] Where: - \( S = -40 \, \text{m} \) (downward displacement) - \( u = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \) (initial vertical velocity) - \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity) Substituting these values into the equation: \[ -40 = 10t - 5t^2 \] Rearranging gives: \[ 5t^2 - 10t - 40 = 0 \] Dividing through by 5: \[ t^2 - 2t - 8 = 0 \] ### Step 2: Solve the quadratic equation We can factor the quadratic equation: \[ (t - 4)(t + 2) = 0 \] This gives us two solutions for \( t \): \[ t = 4 \, \text{s} \quad \text{(valid solution)}, \quad t = -2 \, \text{s} \quad \text{(not valid)} \] Thus, the time of flight \( T1 = 4 \, \text{s} \). ### Step 3: Calculate horizontal velocity (Ux) The horizontal component of the initial velocity is given by: \[ U_x = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \] ### Step 4: Calculate horizontal displacement (D) The horizontal displacement can be calculated using the formula: \[ D = U_x \times T1 \] Substituting the values: \[ D = (10\sqrt{3}) \times 4 = 40\sqrt{3} \, \text{m} \] ### Step 5: Approximate the value of D Using the approximate value of \( \sqrt{3} \approx 1.73 \): \[ D \approx 40 \times 1.73 = 69.2 \, \text{m} \] Thus, the horizontal displacement of the ball from the foot of the tower is approximately: \[ D \approx 70 \, \text{m} \] ### Final Answer The horizontal displacement from the foot of the tower is approximately equal to **70 meters**. ---
To solve the problem, we need to find the horizontal displacement of a ball projected from the top of a 40 m high tower at an initial speed of 20 m/s at an angle of 30 degrees. We will break down the solution into steps. ### Step 1: Determine the time of flight (T1) The total time taken for the ball to hit the ground can be found using the vertical motion equations. The height of the tower is 40 m, and we can use the following kinematic equation: \[ S = ut + \frac{1}{2} a t^2 \] ...
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