Answer:
a) the kinetic energy of the ball at its highest point is 69.58 J
b) its speed when it is 8.11 m below its highest point is 55.97 m/s
Explanation:
Given that;
mass of golf ball m = 46.8 g = 0.0468 kg
initial speed of the ball v₁ = 58.8 m/s
height h = 24.7 m
acceleration due to gravity = 9.8 m/s²
the kinetic energy of the ball at its highest point = ?
from the conservation of energy;
Kinetic energy at the highest point will be;
K.Ei + P.Ei = KEf + PEf
now the Initial potential energy of the ball P.Ei = 0 J
so
1/2mv² + 0 J = KEf + mgh
K.Ef = 1/2mv² - mgh
we substitute
K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]
K.Ef = 80.904 - 11.3284
K.Ef = 69.58 J
Therefore, the kinetic energy of the ball at its highest point is 69.58 J
b) when the ball is 8.11 m below the highest point, speed = ?
so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m
so our velocity will be v₂
also using the principle of energy conservation;
K.Ei + P.Ei = KEh + PEh
1/2mv² + 0 J = 1/2mv₂² + mgh'
1/2mv₂² = 1/2mv² - mgh'
multiply through by 2/m
v₂² = v² - 2gh'
v₂ = √( v² - 2gh' )
we substitute
v₂ = √( (58.8)² - 2×9.8×16.59 )
v₂ = √( 3457.44 - 325.164 )
v₂ = √( 3132.276 )
v₂ = 55.97 m/s
Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s
