Answer:
The ball impact velocity i.e(velocity right before landing) is 6.359 m/s
Explanation:
This problem is related to parabolic motion and can be solved by the following equations:
----------------------(1)
---------(2)
----------------------- (3)
Where:
x = m is the horizontal distance travelled by the golf ball
is the golf ball's initial velocity
is the angle (it was a horizontal shot)
t is the time
y is the final height of the ball
is the initial height of the ball
g is the acceleration due gravity
V is the final velocity of the ball
Step 1: finding t
Let use the equation(2)
s
Substituting (6) in (1):
-------------------(4)
Step 2: Finding :
From equation(4)
m/s (8)
Substituting in (3):
v =42 .01 - 15.3566
V=26.359 m/s
Previous results tell us the speed (v) is given in terms of the coefficient of friction (k) and the radius of the curve (r) as
v = √(kgr)
v = √(0.20·9.8 m/s²·50 m)
= 7√2 m/s ≈ 9.90 m/s
Answer:
mass is lifted 1.8 m. What is the potential energy of the mass 4. A 100 kg