graviyt/maths question

CBORDINO

Member
Hi guys, I have another question, this one is really hard!!!

A woman throws a cricket ball vertically upwards into the air. The ball leaves the woman’s hand 1.8 m above the ground, with a speed of 9.7 m s−1. It rises and then falls back to the ground. By considering the energies involved, calculate the ball’s speed at the point at which it hits the ground.

You may assume that the acceleration due to gravity, g, is 9.8 m s−2 and you may neglect air resistance.

:eek::eek::eek:
 

divethruhaze

Well-known member
Got 11.37 m/s as answer.

Used the kinematics equation:
v^2 = u^2 + 2as

v=final velocity
u=initial velocity
a=acceleration due to gravity
s=displacement from start

Took upward direction to be positive. Therefore the values:
v= to be found
u= 9.7
a= -9.8
s= -1.8

v^2 = (9.7)^2 + (2)*(-9.8)*(-1.8)
v^2 = 129.37
v = 11.37
 
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