JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Kinematics

The maximum speed of a boat in still water is

27 km / h

. Now this boat is moving downstream in a river flowing at

9 km / h

. A man in the boat throws a ball vertically upwards with speed of

10 m / s

. Range of the ball as observed by an observer at rest on the river bank, is

____

cm .
(Take

g = 10 m / s^{2}

)

Answer: 2000

Solution

The boat moves downstream, so its velocity relative to the ground is the sum of its own speed and the current speed:

v = 27 + 9 = 36 km/h = 36 \times \frac{5}{18} = 10 m/s

The ball is thrown vertically upward, relative to the boat, with speed 10 m/s. As seen from the bank, the ball has this horizontal velocity of 10 m/s (carried along with the boat) together with the vertical throw.

Time of flight (rise and fall back to the same height):

T = \frac{2 u}{g} = \frac{2 \times 10}{10} = 2 s

Horizontal range as seen from the bank:

R = v \times T = 10 \times 2 = 20 m = 2000 cm

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Kinematics

Solution

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