Question

A body is thrown horizontally from the top of a tower. It reaches the ground after 4s at an angle 45° to the ground. The velocity of projection is

• A. 9.8 m/s
• B. 19.6 m/s
• C. 29.4 m/s
• D. 39.2 m/s

Answer 1

Answer: (D)

39.2 m/s

Answer 2

The initial velocity has only a horizontal component, $v_{0x} = v_0$, and the initial vertical component is $v_{0y} = 0$. The vertical velocity at time $t$ is given by $v_y = v_{0y} + gt$. With $v_{0y} = 0$, we get $v_y = gt$. The horizontal velocity remains constant throughout the motion, $v_x = v_{0x} = v_0$. The body reaches the ground after $t = 4 \, s$. So, the vertical velocity at impact is $v_y = g \times 4 = 4g$. The velocity of the body when it hits the ground has components $v_x = v_0$ and $v_y = 4g$. The angle $\theta$ that the velocity makes with the ground (which is horizontal) is given by $\tan \theta = \frac{v_y}{v_x}$. We are given that this angle is $45^\circ$. So, $\tan 45^\circ = \frac{4g}{v_0}$. Since $\tan 45^\circ = 1$, we have $1 = \frac{4g}{v_0}$. This implies $v_0 = 4g$. Using the standard value $g = 9.8 \, m/s^2$, we calculate the velocity of projection: $v_0 = 4 \times 9.8 \, m/s = 39.2 \, m/s$.