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Physics Question 82 – AP-EAMCET 2025

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If a body is thrown vertically upwards from the ground with a velocity of 20 m s−1, then its displacement during the last second of upward motion is (Acceleration due to gravity =10 m s−2)

The motion of a body thrown vertically upwards is governed by constant acceleration due to gravity. The velocity decreases as it moves up, becoming zero at the highest point.

Step 1: Calculate the total time of upward motion✦ Active

Identify the initial velocity u=20 m/s and acceleration due to gravity g=10 m/s2. At the maximum height, the final velocity v=0. Use the first equation of motion, v=u+at, with a=−g for upward motion.

0=u−gT ⇒T=ug

Substitute the given values:

T=20 m/s10 m/s2=2 s

The body takes 2 s to reach its maximum height.

💡 Teacher's Secret Hint

Remember that acceleration due to gravity acts downwards, so when the body is moving upwards, its velocity decreases, and we use a=−g.

Step 2: Identify the 'last second' of upward motion○ Expand

Since the total time of upward motion is 2 s, the 'last second' refers to the time interval from t=1 s to t=2 s, which is the 2nd second of its journey.

Step 3: Calculate the displacement during the last second○ Expand

We can use the formula for displacement in the n-th second:

sn=u+a2(2n−1)

Here, u=20 m/s, a=−g=−10 m/s2, and n=2 (for the 2nd second).

s2=20+−102(2×2−1)s2=20−5(4−1)s2=20−5(3)s2=20−15s2=5 m

Alternatively, calculate the displacement in the first second (s1) and the displacement in the first two seconds (s2). The displacement in the last second is s2−s1.

s1=ut+12at2=(20)(1)+12(−10)(1)2=20−5=15 ms2=ut+12at2=(20)(2)+12(−10)(2)2=40−20=20 mslast second=s2−s1=20 m−15 m=5 m
💡 Teacher's Secret Hint

A common mistake is to calculate the displacement from the start of the motion. The question asks for displacement *during* the last second, which is the distance covered in that specific one-second interval.

Step 4: State the final answer○ Expand

The displacement during the last second of upward motion is 5 m.

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