Question 11: Velocity of Projection from s-t Graph
Solution
The graph plots the distance $s$ of a ball from a fixed point $P$ versus time. Since the ball is thrown vertically upwards, its trajectory is a straight vertical line. The fixed point $P$ is at a perpendicular distance $d$ from this line.
1. Analyzing the Geometry:
From the graph:
- Minimum distance $s_3 = 6$ m. This occurs when the ball is at the same vertical level as point $P$. Thus, horizontal separation $d = 6$ m.
- Initial distance $s_1 = 6\sqrt{5}$ m. Using Pythagoras theorem, the vertical distance of the launch point below level $P$ is: $$ y_1 = \sqrt{s_1^2 – d^2} = \sqrt{(6\sqrt{5})^2 – 6^2} = \sqrt{180 – 36} = \sqrt{144} = 12 \text{ m} $$
- Distance at the second peak $s_2 = 10$ m. In the “W” shaped graph for vertical motion, the local maximum between two minima corresponds to the maximum height of the ball (apogee).
Vertical distance of max height above level $P$ is: $$ y_2 = \sqrt{s_2^2 – d^2} = \sqrt{10^2 – 6^2} = \sqrt{100 – 36} = \sqrt{64} = 8 \text{ m} $$
2. Kinematics:
The ball travels from the launch point (12 m below $P$) to the maximum height (8 m above $P$).
Total vertical displacement to maximum height:
$$ H_{\text{total}} = y_1 + y_2 = 12 + 8 = 20 \text{ m} $$
At maximum height, final velocity $v = 0$. Using the equation of motion $v^2 = u^2 – 2gH$: $$ 0 = u^2 – 2(10)(20) $$ $$ u^2 = 400 $$ $$ u = 20 \text{ m/s} $$
Answer: 20 m/s