Question

If a ball is thrown horizontally with a velocity of $10\text{ ms}^{-1}$ from the tower of height 45 m, then it strikes the ground at a distance of $(g=10\text{ ms}^{-2})$

• A. 25 m
• B. 40 m
• C. 20 m
• D. 30 m
• E. 45 m

Hint:

Logic Tip: The formula for time of flight in a pure horizontal projection is always $t = \sqrt{\frac{2h}{g}}$. Memorizing this allows you to skip the kinematic setup entirely: $t = \sqrt{90/10} = \sqrt{9} = 3$ seconds. Then multiply by $u_x$ immediately.

Answer 1

The Correct Option is D

Concept:
In horizontal projectile motion, the horizontal and vertical motions are independent. Vertical motion dictates the time of flight $t$ via the equation $h = u_yt + \frac{1}{2}gt^2$. Horizontal motion dictates the range $R$ via the equation $R = u_x \times t$.
Step 1: Identify the initial conditions.
Horizontal initial velocity: $u_x = 10\text{ ms}^{-1}$ Vertical initial velocity: $u_y = 0\text{ ms}^{-1}$ (since it is thrown perfectly horizontally) Height of the tower: $h = 45\text{ m}$ Acceleration due to gravity: $g = 10\text{ ms}^{-2}$
Step 2: Calculate the time of flight.
Use the vertical kinematic equation: $$h = u_y t + \frac{1}{2} g t^2$$ Substitute the known values: $$45 = (0)t + \frac{1}{2}(10)t^2$$ $$45 = 5t^2$$ $$t^2 = 9$$ $$t = 3\text{ seconds}$$
Step 3: Calculate the horizontal range.
The horizontal distance (Range) is simply the horizontal velocity multiplied by the time of flight, as there is no horizontal acceleration. $$R = u_x \times t$$ $$R = 10 \times 3$$ $$R = 30\text{ m}$$