2014 Physics WAEC (WASSCE) Essay Past Questions (Paper 2)

Question 1

A particle is projected horizontally at 10 ms^-1 from the top of a tower 20 m high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g = 10 ms^-2 ]

Solution

Horizontal distance, R = \(\scriptsize U_x \: \times \: t\)

• U_x = 10 ms^-1
• t is the time taken to reach the ground

To calculate t,

At maximum height, H, the velocity = 0.

The initial velocity is zero when an object is released from a height, H.

When an object strikes the ground, H = 0,

If an object falls from a height, H, U_y = 0

H = \(\scriptsize U_yt \: + \:  \normalsize \frac{1}{2} \scriptsize gt^2\)

H = \(\scriptsize 0 \: + \: \normalsize \frac{1}{2} \scriptsize gt^2\)

t² =  \(\frac{2H}{g}\)

t = \(\sqrt{ \frac{2H}{g}}\)

∴ R = \(\scriptsize U_x  \: \times \:  \sqrt{ \normalsize\frac{2H}{g}}\)

R = \(\scriptsize 10 \: \times \:  \sqrt{\normalsize \frac{2 \: \times \: 20}{10}}\)

R = \(\scriptsize 10\: \times \: 2\)

R = 20 m

Question 2

A tennis ball projected at an angle θ attains a range R = 78 m. If the velocity imparted to the ball by the racket is 30 ms^-1 , calculate θ. [g = 10 ms^-2 ]

Solution 

Recall that:-

R = \(\frac{u^2sin2 \theta}{g}\)

Information given:- Range, R= 78 m, acceleration due to gravity,g = 10 ms^-2 ,

θ = ?, U = 30 ms^-1

R = \(\frac{30^2 sin2 \theta}{10}\)

sin2θ =  \(\frac{780}{900}\)

sin2θ = 0.8667

2θ = sin^-1 0.8667 = 60°

θ = \(\frac{60^o}{2}\)

θ = 30°