Question

A projectile is thrown with speed 20 m/s from ground at angle of 37° with horizontal. At highest point, projectile is broken into two pieces. One piece (one third of original piece) falls vertically and other hits ground x m away from point of projection, then x is

• A. 40 m
• B. 50 m
• C. 60 m
• D. 48 m

Answer 1

Answer: (D)

48 m

Answer 2

The problem involves projectile motion followed by an explosion and conservation of momentum.

1. Analyze the initial projectile motion up to the highest point: Given:

• Initial speed, $u = 20 \, \text{m/s}$
• Angle of projection, $\theta = 37^\circ$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$ (standard assumption)

We use the approximate values: $\sin 37^\circ \approx 3/5 = 0.6$ and $\cos 37^\circ \approx 4/5 = 0.8$.

• Initial horizontal velocity component: $u_x = u \cos \theta = 20 \times \cos 37^\circ = 20 \times (4/5) = 16 \, \text{m/s}$
• Initial vertical velocity component: $u_y = u \sin \theta = 20 \times \sin 37^\circ = 20 \times (3/5) = 12 \, \text{m/s}$

At the highest point of the trajectory:

• Vertical velocity becomes zero ($v_y = 0$).

• Horizontal velocity remains constant ($v_x = u_x = 16 \, \text{m/s}$).

• Time taken to reach the highest point ($t_H$): Using $v_y = u_y - g t_H$: $0 = 12 - 10 t_H \implies t_H = 1.2 \, \text{s}$

• Horizontal distance covered to the highest point ($R_H$): $R_H = u_x t_H = 16 \times 1.2 = 19.2 \, \text{m}$

• Maximum height ($H$): Using $H = u_y t_H - \frac{1}{2} g t_H^2$: $H = 12 \times 1.2 - \frac{1}{2} \times 10 \times (1.2)^2 = 14.4 - 5 \times 1.44 = 14.4 - 7.2 = 7.2 \, \text{m}$

2. Apply conservation of momentum at the highest point: Let the original mass of the projectile be $M$. At the highest point, just before breaking, the projectile's velocity is $\vec{V} = 16 \hat{i} \, \text{m/s}$ (since $v_y = 0$).

The projectile breaks into two pieces:

• Piece 1: mass $m_1 = M/3$. It "falls vertically". This implies its horizontal velocity becomes zero ($v_{1x} = 0$). Also, for simplicity and common interpretation in such problems, we assume its initial vertical velocity immediately after the explosion is zero ($v_{1y} = 0$).
• Piece 2: mass $m_2 = M - M/3 = 2M/3$. Let its velocity be $\vec{v_2} = v_{2x} \hat{i} + v_{2y} \hat{j}$.

Apply conservation of linear momentum:

• Horizontal direction: (No external horizontal forces during explosion) $M V_x = m_1 v_{1x} + m_2 v_{2x}$ $M (16) = (M/3)(0) + (2M/3) v_{2x}$ $16 = (2/3) v_{2x}$ $v_{2x} = 16 \times (3/2) = 24 \, \text{m/s}$

• Vertical direction: (No external vertical forces during explosion, and initial vertical velocity is zero) $M V_y = m_1 v_{1y} + m_2 v_{2y}$ $M (0) = (M/3)(0) + (2M/3) v_{2y}$ $0 = (2M/3) v_{2y} \implies v_{2y} = 0$

So, immediately after the explosion, the second piece has a velocity of $\vec{v_2} = 24 \hat{i} \, \text{m/s}$. It is at a height of $H = 7.2 \, \text{m}$ above the ground.

3. Analyze the motion of the second piece until it hits the ground: The second piece starts from height $H = 7.2 \, \text{m}$ with initial horizontal velocity $v_{2x} = 24 \, \text{m/s}$ and initial vertical velocity $v_{2y} = 0$.

• Time taken for the second piece to fall to the ground ($t_f$): Using $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$: $0 = 7.2 + 0 \cdot t_f - \frac{1}{2} \times 10 \times t_f^2$ $0 = 7.2 - 5 t_f^2$ $5 t_f^2 = 7.2$ $t_f^2 = 7.2 / 5 = 1.44$ $t_f = \sqrt{1.44} = 1.2 \, \text{s}$

• Additional horizontal distance covered by the second piece during $t_f$ ($R_2$): $R_2 = v_{2x} t_f = 24 \times 1.2 = 28.8 \, \text{m}$

4. Calculate the total distance 'x': The total distance 'x' from the point of projection where the second piece hits the ground is the sum of the horizontal distance to the highest point and the additional horizontal distance covered by the second piece. $x = R_H + R_2 = 19.2 \, \text{m} + 28.8 \, \text{m} = 48.0 \, \text{m}$

The final answer is $\boxed{\text{48 m}}$.

Explanation of the solution:

1. Calculate the horizontal distance ($R_H$) and height ($H$) reached by the projectile at its highest point using initial velocity components ($u_x = u \cos 37^\circ$, $u_y = u \sin 37^\circ$) and time to reach peak ($t_H = u_y/g$).
2. At the highest point, apply conservation of horizontal momentum to find the horizontal velocity of the second piece ($v_{2x}$). Assume the first piece falling vertically means its horizontal velocity is zero and its vertical velocity is also zero immediately after the explosion. This implies the second piece also has zero vertical velocity immediately after the explosion.
3. Calculate the time ($t_f$) it takes for the second piece to fall from height $H$ to the ground, starting with zero initial vertical velocity.
4. Calculate the additional horizontal distance ($R_2$) covered by the second piece during this fall time, using its horizontal velocity ($v_{2x}$).
5. The total distance 'x' is the sum of $R_H$ and $R_2$.