9702/Oct Nov/11/2016/Q6

Question 6

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into

the air and onto a track that is vertically 2.0 m lower.

The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.

What is the horizontal velocity v of the cyclist at the end of the higher track?

A      6.3 ms^–1 

B      9.4 ms^–1 

C      9.9 ms^–1 

D      15 ms^–1

Solution:

There are important points to note in this question:

The horizontal velocity v is used to calculate the horizontal distance

The time to reach the maximum height is the time to travel the horizontal distance

At maximum height u = 0

Using H = ut + 1/2gt^2

2 = 0 + 1/2×9.81xt^2

(t=0.6395s)

Horizontal distance = horizontal velocity(v) x time(t)

6 = 0.6395v

V = 9.4ms^-2 (B is the correct option)

Ans: B

Reference PYQ - Oct/Nov 2016 Paper 11 Q6