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
C and d have the same amount of protons and electrons
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
Answer:
Using g = 9.8: 1.02 kg, Using g = 10: 1 kg
Explanation:
E = mgh
20 = m(9.8)(3 - 1)
20 = 9.8m(2)
20 = 19.6m
m = 1.02 kg
I'm now assuming you may be using a g constant of 10, thus the close integer result, in which case the mass would be exactly 1 kilogram.