Question

One second after projection, the horizontal and vertical velocities of a projectile are found to be equal and after one more second, the motion of the projectile is along the horizontal. The horizontal range of the projectile is \[ (g=10~\text{m s}^{-2}) \]

• A. \(10~\text{m}\)
• B. \(20~\text{m}\)
• C. \(30~\text{m}\)
• D. \(40~\text{m}\)

Hint:

Projectile formulas: \[ \boxed{ T=\frac{2u_y}{g}, \qquad R=u_xT } \] Horizontal motion has \[ \boxed{v_y=0.} \]

Answer 1

The Correct Option is D

Step 1: Use the second condition. After one more second (i.e., at \(t=2\) s), the projectile moves horizontally. Hence, \[ v_y=u_y-gt=0 \] \[ u_y=2g=20~\text{m s}^{-1}. \]

Step 2: Use the first condition. At \(t=1\) s, \[ v_x=v_y. \] Now, \[ v_x=u_x, \] and \[ v_y=u_y-g =20-10 =10~\text{m s}^{-1}. \] Therefore, \[ u_x=10~\text{m s}^{-1}. \]

Step 3: Calculate the range. Time of flight: \[ T=\frac{2u_y}{g} =\frac{2\times20}{10} =4~\text{s}. \] Range: \[ R=u_xT =10\times4 =40~\text{m}. \] Thus, \[ \boxed{R=40~\text{m}} \] Hence, \[ \boxed{(D)} \] is the correct answer.