Question

If a ball is thrown vertically upwards with speed $u,$ the distance covered during the last $'t'$ seconds of its ascent is

• A. $ut$
• B. $\frac{1}{2}gt^2$
• C. $ut-\frac{1}{2}gt^2$
• D. $(u + gt )\, t$

Answer 1

The Correct Option is B

To solve the problem of finding the distance covered during the last $t$ seconds of ascent when a ball is thrown vertically upwards with speed $u$, we need to understand the physics involved in the motion under uniform acceleration due to gravity.

Concept: When the ball is thrown upwards, it will continue to ascend until its velocity becomes zero at the maximum height. The acceleration due to gravity $(g)$ acts downwards and affects the motion.

Now, consider the motion equations during the ascent:

• The velocity of the ball at any time $t$ is given by: \( v = u - gt \)
• The total time taken to reach the maximum height is when $v = 0$, so: \( 0 = u - gt' \Rightarrow t' = \frac{u}{g} \)

Distance Calculated: To find the distance during the last $t$ seconds, we use the kinematic equation for distance: \( h = ut' - \frac{1}{2}g{t'}^2 \)

For the last $t$ seconds of the ascent, the initial velocity during these last seconds can be considered as \( v_o = g(t'-t) \) because: \( v_o = v + gt = 0 + g(t'-t) \)

The distance $d$ covered in the last $t$ seconds can be derived from: \( d = v_ot - \frac{1}{2}gt^2 \), where \( v_o = gt \).

Therefore, substituting \( v_o \) into the distance equation gives: \( d = gt\cdot t - \frac{1}{2}gt^2 = \frac{1}{2}gt^2 \)

Conclusion: Thus, the distance covered during the last $t$ seconds of its ascent is \(\frac{1}{2}gt^2\).