Agolf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. the drive lands 130 m away on a flat fairway. the acceleration of gravity is 9.8 m/s 2 . if the golf club and ball are in contact for 6.3 ms, what is the average force of impact? neglect air resistance. answer in units of n.

The force is calculated as 338.66 N

Explanation:

We know that force is given by

We know that range of a projectile is given by

it is given that R=130 m applying values in the above equation we get

Thus the force is obtained as

Thus force equals

522.84 N

Explanation:

m = 59.2 g = 59.2 x 10^-3 kg

θ = 50.5

R = 130 m

g = 9.8 m/s^2

t = 6.3 m s = 6.3 x 10^-3 s

Let u be the velocity of projection

Use the formula for the range

130 = u^2 x Sin 2(50.5) / 9.8

u = 36.06 m/s

The vector form of the initial velocity

u = 36.05 (Cos 50.5 i + Sin 50.5 j)

u = 22.93 i + 27.82 j

The velocity of the body as it strikes with the surface is same but the direction is downward.

v = 22.93 i - 27.82 j

Change in momentum, Δp = m (v - u)

Δp = 59.2 x 10^-3 x (22.93 i - 27.82 j - 22.93 i - 27.82 j)

Δp = 59.2 x 10^-3 x ( - 55.64 j)

Δp = - 3.294 j

Δp = 3.24

Force = Rate of change in momentum

F = Δp / Δt = 3.294 / (6.3 x 10^-3) = 522.84 N

b would be the answer

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