A ball rolls along a 1.0m high horizontal table and then falls off of the edge of the table with a horizontal velocity of 4.0ms-1 Calculate:

a. time taken for the ball to reach the floor after it leaves the table (2 mark)

b.the horizontal distance travelled by the ball during its fall (2 mark)

2) A bullet is fired horizontally from a gun held 1.4m above the ground. If the bullet leaves the gun with a velocity of 300ms-1 at what distance from the gun will the bullet strike the ground? (3 mark)

a. To calculate the time taken for the ball to reach the floor after it leaves the table, we can use the equation for vertical motion:

h = (1/2)gt^2

Where:
h = height (1.0m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Rearranging the equation to solve for time:

t = sqrt(2h/g)

Substituting the given values:

t = sqrt(2 * 1.0 / 9.8)
t ≈ sqrt(0.204)
t ≈ 0.452 s

Therefore, the time taken for the ball to reach the floor after it leaves the table is approximately 0.452 seconds.

b. The horizontal distance traveled by the ball during its fall can be calculated using the equation:

d = vt

Where:
d = horizontal distance
v = horizontal velocity (4.0 m/s in this case)
t = time (0.452 s from part a)

Substituting the given values:

d = 4.0 * 0.452
d ≈ 1.808 m

Therefore, the horizontal distance traveled by the ball during its fall is approximately 1.808 meters.

2) To calculate the distance from the gun at which the bullet strikes the ground, we need to consider the horizontal motion of the bullet. Since the bullet is fired horizontally, the initial vertical velocity (Vy) is zero.

Using the equation for vertical motion:

h = (1/2)gt^2

Where:
h = height (1.4 m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

We can solve for t:

t = sqrt(2h/g)

Substituting the given values:

t = sqrt(2 * 1.4 / 9.8)
t ≈ sqrt(0.2857)
t ≈ 0.5355 s

Now that we have the time, we can calculate the horizontal distance (d) using:

d = vt

Where:
v = horizontal velocity (300 m/s in this case)
t = time (0.5355 s from above)

Substituting the given values:

d = 300 * 0.5355
d ≈ 160.65 m

Therefore, the bullet will strike the ground at a distance of approximately 160.65 meters from the gun.

To answer these questions, we need to use the equations of motion. Let's break down each question step-by-step:

For Question 1:
a. To calculate the time taken for the ball to reach the floor, we need to find the time it takes for the ball to fall after leaving the table. We can use the equation of motion in the vertical direction:

h = ut + (1/2)gt^2

Where h is the height, u is the initial vertical velocity (0 m/s since the ball is not given an initial vertical velocity), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.

Since the height is 1.0 m and the initial vertical velocity is 0 m/s, we can rearrange the equation to solve for time:

t = sqrt(2h/g)

Plugging in the values, we get:

t = sqrt(2 * 1.0 / 9.8) ≈ 0.45 s

b. To calculate the horizontal distance traveled by the ball during its fall, we can use the horizontal velocity and the time calculated in part (a). The equation to calculate distance is:

d = vt

Where d is the horizontal distance, v is the horizontal velocity, and t is the time.

Plugging in the values, we get:

d = 4.0 m/s * 0.45 s ≈ 1.8 m

So, the time taken for the ball to reach the floor after leaving the table is approximately 0.45 seconds, and the horizontal distance traveled by the ball during its fall is approximately 1.8 meters.

For Question 2:
To calculate the distance from the gun where the bullet will strike the ground, we can break down the motion of the bullet into horizontal and vertical components.

Since the bullet is fired horizontally, the initial vertical velocity is 0 m/s. The vertical motion is influenced by gravity, so we can use the equation:

h = ut + (1/2)gt^2

Again, h is the height (1.4 m), u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.

Solving for time, we get:

t = sqrt(2h/g)

Plugging in the values, we get:

t = sqrt(2 * 1.4 / 9.8) ≈ 0.53 s

Since the bullet was fired horizontally, the horizontal distance it travels can be calculated as:

d = vt

Where d is the horizontal distance, v is the horizontal velocity (300 m/s), and t is the time.

Plugging in the values, we get:

d = 300 m/s * 0.53 s ≈ 159 m

So, the bullet will strike the ground at a horizontal distance of approximately 159 meters from the gun.

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