First slide

Projection Under uniform Acceleration

Question

A body is projected horizontally from the top of a tower with initial velocity 18 ms^-1. It hits the ground at angle 45°. What is the vertical component of velocity when strikes the ground?

Moderate

A

9 ms-1
B

$9 \sqrt{2}$ ms-¹
C

18 ms-¹
D

$18 \sqrt{2}$ ms^-1

Solution

$vcos 45 ° = u = 18 {ms}^{- 1} \Rightarrow v = 18 \sqrt{2} {ms}^{- 1} Vertical component$
$y \sin 45 ° = 18 \sqrt{2} \times \frac{1}{\sqrt{2}} = 18 {ms}^{- 1} ∣$

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