First slide

Projection Under uniform Acceleration

Question

A grasshopper can jump a maximum distance 1.6 m. It spends negligible time on the ground. How far can it go in 10 s?

Easy

A

$5 \sqrt{2} m$
B

$10 \sqrt{2} m$
C

$20 \sqrt{2} m$
D

$40 \sqrt{2} m$

Solution

Here angle of projection = 45°
$R_{max} = 1 .6 = \frac{u^{2}}{g} = \frac{u^{2}}{10} or u = 4 {ms}^{- 1} Hence, the distance covered in 10 s is horizontal speed x time = u \cos 45 ° \times time = 4 \times \frac{1}{\sqrt{2}} \times 10 = 20 \sqrt{2} m ∣$

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