The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an angle of 600 with the horizontal. Calculate the maximum height achieved by the ball.

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

Hence the maximum height achieved by the ball is 61.22m

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever