Question

You throw a rock horizontally off a cliff with a speed of 20 m/s. after two seconds, the magnitude of the velocity of the rock is closest to

Answer 1

So the magnitude of the resultant vector of the vertical and horizontal together is a right triangle with 20 on one side and 19.62 going down. 
A^2 + B^2 = C^2 
20^2 + 19.62^2 = C^2 
400 + 385 = C^2 
C = 28.0 m/s magnitude 

Answer 2

We have that the magnitude of the velocity of the rock is closest to

V=28m/s

From the question we are told

You throw a rock horizontally off a cliff with a speed of 20 m/s. after two seconds, the magnitude of the velocity of the rock is closest to

• 28
• 37
• 20
• 40

Generally the Newtons equation for Motion mathematically given as

v=u+at

Therefore

For Vertical Axis

$V_v=0+9.8*2\\\\V_v=19.61$

Therefore

the magnitude of the velocity

$V=|sqrt{V_v^2+V_h^2}\\\\V=|sqrt{20^2^2+19.61^2}\\\\V=28m/s$

Therefore

The magnitude of the velocity of the rock is closest to

V=28m/s

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