Motion Under Gravity Numerical: Calculating Ball’s Speed at Point C

Question:

A ball is thrown vertically upwards with an initial velocity of 5 ms^-1 from point A as shown below. C is a point 10 m vertically below the point A. The the speed of the ball at point C will be(take g=10 ms^-2 and neglect air resistance)

• a) 7.5 ms^-1
• b) 10 ms^-1
• c) 15 ms^-1
• d) 17.5 ms^-1

Solution:

Initial velocity u= +5 ms^-1 (From the point A, upwards)

Final velocity v = 0 (At the maximum point the velocity is zero)

Acceleration due to gravity a=-g=-10 ms^-2

Displacement s= y-y_o

Applying the third equation of motion

$${v^2=u^2+2 a s}$$

$${0 =5^2 + 2 (-10) s}$$

$${s=\frac {25} {20}}$$

$${s=1.25 m}$$

s= y-y_o = 1.25 m . Therefore y=1.25+y_o =1.25 +10 =11.25 m

(ii) Speed of the ball just before hitting the ground

$${v=\sqrt 2 g h}$$

$${v=\sqrt {(2)(10)(11.25)}}$$

$${v=\sqrt 225}$$

$${v= 15 \frac{m}{s}}$$

The correct option is (c) 15 ms^-1