A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the
1 answer:
Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a)
b)
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -<em>g</em>. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: .
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.
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Explanation:
1.
We use the equation
h = , where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Answer:
Explanation:
Using kinematics equations:
Use due to condition of distance traveled.
Solving second equation for time, there are two solutions. t=0 and
Use the expression in the first equation to have
Using trigonometric identities, you have the answer of the distance.
By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.