A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45° ramp launching the ball into the hole which is d = 1.60 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, y = 0.690 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface and the acceleration due to gravity is 9.81 m/s2.

4.74m/s

5.41

To calculate the speed at which the golfer must hit the ball, we can use the principles of projectile motion. We can break down the motion into horizontal and vertical components.

1. Vertical motion:
The ball starts at a height y = 0.690 m and lands in a hole that is at the same height. The vertical distance traveled by the ball is 0 m. The only force acting on the ball in the vertical direction is gravity, which accelerates the ball downwards with an acceleration of 9.81 m/s².

Since the vertical distance traveled is 0 m and the initial vertical velocity is zero, we can use the following equation:

y = y₀ + v₀y*t + (1/2)*a*t²

Where:
y₀ = initial vertical position = 0.69 m
v₀y = initial vertical velocity = 0 m/s
a = acceleration due to gravity = -9.81 m/s² (negative because it acts downward)
t = time taken to reach the hole

Plugging in the values, the equation becomes:

0 = 0.69 + 0 - (1/2)*9.81*t²

Simplifying:

0 = 0.69 - 4.905*t²

Rearranging the equation:

4.905*t² = 0.69

t² = 0.69 / 4.905

t ≈ 0.316 s

2. Horizontal motion:
The horizontal distance the ball needs to travel is d = 1.60 m. The horizontal motion is not influenced by gravity since there are no horizontal forces acting on the ball. The only force acting on the ball horizontally is provided by the golfer'.

The horizontal distance traveled by the ball is given by:

d = v₀x*t

Where:
v₀x = initial horizontal velocity
t = time taken to reach the hole

Rearranging the equation:

v₀x = d / t

Plugging in the values:

v₀x = 1.60 m / 0.316 s

v₀x ≈ 5.06 m/s

3. Overall speed:
The overall speed at which the golfer must hit the ball is given by the combined horizontal and vertical velocities. Since there is no vertical velocity initially, the overall speed is equal to the horizontal speed:

Overall speed = v₀x ≈ 5.06 m/s

Therefore, the golfer must hit the ball at a speed of approximately 5.06 m/s.

To find the speed at which the golfer must hit the ball, we can apply the principle of conservation of energy. The initial energy of the ball when it is hit is converted into potential energy as it rises up the ramp and then into kinetic energy as it moves towards the hole.

1. Let's calculate the potential energy of the ball at the top of the ramp. The potential energy equation is given by PE = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the reference point (y = 0 in this case).

Since the height of the ball is given as y = 0.690 m, the potential energy at the top of the ramp is PE = m * 9.81 * 0.690.

2. Next, let's calculate the distance along the ramp that the ball travels. The ball moves along the ramp at an angle of 45°, which means it travels a horizontal distance equal to the hypotenuse of a right triangle with sides of length 1.60 m.

Using the Pythagorean theorem, the horizontal distance traveled along the ramp is given by d = 1.60 m.

3. Now, let's calculate the potential energy of the ball just before it lands in the hole. The potential energy at this point is zero since it is at the same height as the reference point (y = 0).

4. Finally, let's equate the initial potential energy to the final potential energy and solve for the kinetic energy of the ball. Considering that the total energy is conserved (assuming no energy losses due to friction), we have:

PE(initial) = PE(final) + KE

Substituting the values calculated in steps 1 and 3:

m * 9.81 * 0.690 = 0 + 0.5 * m * v^2

Simplifying the equation and solving for v (the speed of the ball):

v = √((2 * 9.81 * 0.690))

Plugging in the values and calculating:

v ≈ 7.23 m/s

Therefore, the golfer must hit the ball with a speed of approximately 7.23 m/s to land it in the hole.