- #1
imy786
- 322
- 0
Homework Statement
The distance from the Cricketer to the wall is 70 metres across level ground.
The ball reaches its maximum height of 30 metres as it passes directly above the wall
Both the height and the distance to the wall are measured from the point at which the ball is hit by the cricketer.
-------------------------------------------------------------------------------------------
1.Calculate the vertical component of the ball's velocity at the point of hit.
2.Calculate the time taken for the ball to reach the highest point.
3.Hence find the horizontal component of the ball’s initial velocity
4.Use the answers to parts (1) and (3) to calculate the magnitude and direction
of the ball’s initial velocity and of its velocity at the highest point.
The cricketer stands back to a distance of 80 metres. When hitting now, the ball only just clears the top of the wall. Calculate the height of the wall.
Homework Equations
Will be completing 1 part at a time.
The Attempt at a Solution
1.Calculate the vertical component of the ball's velocity at the point of release.
v^2 = u^2 + 2as
u = ucosθ
v=0
s=70
a=-9.8
v^2 = u^2 + 2as
0= (ucosθ) ^2 - (19.6 x 70)
1372 = (ucosθ) ^2
37.04 = ucosθ
37.0 m/s is the vertical component of balls velocity. (is this correct for part 1).