First slide

Projection Under uniform Acceleration

Question

A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of 5 ${ms}^{- l}$ . Ignore air resistance and take g = l0 ${ms}^{- 2}$ . Th" speed with which he rouches the peak -B is:

Moderate

A

2.0 ${ms}^{- 1}$
B

15 ${ms}^{- 1}$
C

25 ${ms}^{- 1}$
D

20 ${ms}^{- 1}$

Solution

Speed in horizontal direction remains constant during whole journey because there is no acceleration in this direction.
So, $v_{h} = 6 {ms}^{- 1}$
In vertical direction:
Loss of gravitation potential energy = gain in KE
$i . e ., mgh = \frac{1}{2} {mv}_{V}^{2}$
${v_{V}}^{2} = 2 gh = 2 \times 10 \times (70 - 60) = 200$
Hence, the speed with which he touches the cliff B is
$v = \sqrt{{v_{B}}^{2} + {v_{V}}^{2}} = \sqrt{25 + 200} = \sqrt{225} = 15 {ms}^{- 1}$

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