Question

A bullet is projected upwards from the top of a tower of height 90 m with the velocity 30 m/s making an angle 30° with the horizontal. The time taken by it to reach the ground is _______ $(g = 10m/s^2)$.

• A. 2 sec
• B. 3 sec
• C. 24 sec
• D. 6 sec

Answer 1

Answer: (D)

6 sec

Answer 2

The bullet's motion is projectile motion. We analyze the vertical motion using the kinematic equation $S_y = u_y t + \frac{1}{2} a_y t^2$. Given the initial vertical velocity ($u \sin \theta = 15$ m/s), the vertical displacement to the ground ($-90$ m, taking the tower top as origin), and the acceleration due to gravity ($-10$ m/s$^2$), we set up a quadratic equation $5t^2 - 15t - 90 = 0$. Solving this equation yields $t=6$ s and $t=-3$ s. Since time must be positive, the time taken is 6 seconds.