Kinematics

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Kinematics — solved practice questions

8 JEE Physics questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   A ball is thrown vertically upward with an initial speed of 20 m/s (take g = 10 m/s^2). What is the maximum height it reaches?

   • A10 m
   • B15 m
   • C20 m
   • D40 m
   Show answer & solution

   Correct answer: (C) 20 m

   At the top v = 0, so using v^2 = u^2 - 2gh gives h = u^2/(2g) = 400/20 = 20 m.

2. Q2medium

   A car starting from rest moves with uniform acceleration. If it covers a distance s in the first t seconds, the distance covered in the next t seconds is:

   • A2s
   • B3s
   • C4s
   • Ds
   Show answer & solution

   Correct answer: (B) 3s

   Distance in time t is s = (1/2)a t^2; in time 2t it is (1/2)a(2t)^2 = 4s, so distance in the second interval = 4s - s = 3s.

3. Q3medium

   The position of a particle is x = 4t - 2t^2 (SI units). At the instant its velocity becomes zero, its position is:

   • A2 m
   • B0 m
   • C4 m
   • D1 m
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   Correct answer: (A) 2 m

   v = dx/dt = 4 - 4t = 0 gives t = 1 s; then x = 4(1) - 2(1)^2 = 2 m.

4. Q4easy

   A man can swim at 5 m/s in still water. He wants to cross a river 100 m wide flowing at 3 m/s. If he heads straight across (perpendicular to the bank), the time to cross is:

   • A25 s
   • B12.5 s
   • C33.3 s
   • D20 s
   Show answer & solution

   Correct answer: (D) 20 s

   The crossing time depends only on the perpendicular component of velocity: t = width / (swim speed) = 100/5 = 20 s. The current does not change this time, only the drift.

5. Q5medium

   A projectile is fired with speed u at angle theta with the horizontal. The ratio of its maximum height H to horizontal range R is:

   • Atan(theta)
   • Btan(theta)/2
   • Ctan(theta)/4
   • D4 tan(theta)
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   Correct answer: (C) tan(theta)/4

   H = u^2 sin^2(theta)/(2g) and R = u^2 sin(2theta)/g = 2u^2 sin(theta)cos(theta)/g, so H/R = tan(theta)/4.

6. Q6medium

   Rain is falling vertically at 4 m/s. A man walks horizontally at 3 m/s. At what angle from the vertical must he hold his umbrella to stay dry?

   • A53 degrees
   • B37 degrees
   • C45 degrees
   • D30 degrees
   Show answer & solution

   Correct answer: (B) 37 degrees

   In the man's frame the rain has a horizontal component 3 m/s and vertical 4 m/s, so the umbrella must tilt forward at angle theta from vertical with tan(theta) = 3/4, i.e. about 37 degrees.

7. Q7easy

   A body is projected horizontally with speed 10 m/s from a height of 20 m (g = 10 m/s^2). The horizontal distance it travels before hitting the ground is:

   • A10 m
   • B40 m
   • C30 m
   • D20 m
   Show answer & solution

   Correct answer: (D) 20 m

   Time to fall: t = sqrt(2h/g) = sqrt(4) = 2 s; horizontal range = 10 * 2 = 20 m.

8. Q8medium

   Two balls are projected from the same point with the same speed at angles 30 degrees and 60 degrees to the horizontal. Which of the following is the same for both?

   • AHorizontal range
   • BMaximum height
   • CTime of flight
   • DVelocity at the highest point
   Show answer & solution

   Correct answer: (A) Horizontal range

   Since sin(60) = sin(120), the range R = u^2 sin(2theta)/g is equal for complementary angles 30 and 60 degrees. Maximum heights and times of flight differ.

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