A ball rolls off a platform that is 3 meters above the ground. The ball's horizontal velocity as it leaves the platform is 4.5 m/s.
(a) How much time does it take for the ball to hit the ground? (See the example box, use g = 10 m/s2.)
(b) How far from the base of the platform does the ball hit the ground?
a) Well, it seems like gravity is playing a game called "catch the ball" with us here. To help us solve this mystery, we'll use the good ol' kinematic equation for vertical motion:
y = y₀ + v₀t + (1/2)gt²
Where:
y is the vertical distance traveled (which in our case is 3 meters down),
y₀ is the initial vertical position (which is 0 since the ball starts at the top),
v₀ is the initial vertical velocity (which is 0 because the ball only has horizontal velocity),
t is the time we're trying to find, and
g is the acceleration due to gravity (a thrilling 10 m/s²).
Simplifying this equation, we get:
3 = 0 + 0t + (1/2)(10)t²
Now, let's solve this quadratic equation and unlock the secret hiding in t!
1/2(10)t² = 3
5t² = 6
t² = 6/5
t ≈ √(6/5)
t ≈ √6 / √5
t ≈ √(6/5) seconds
b) Ah, the horizontal motion of the ball is as exciting as watching paint dry. Let's use the horizontal velocity to tell us how far the ball will go before crashing down to Earth.
d = v₀t
Where:
d is the horizontal distance we're interested in,
v₀ is the initial horizontal velocity (4.5 m/s),
and t is the time we just calculated (√(6/5) seconds).
Plugging in the values, we get:
d = (4.5)√(6/5)
d ≈ 4.5√1.2
d ≈ 4.5 * 1.095
d ≈ 4.9275 meters
So, the ball will hit the ground approximately 4.9275 meters away from the base of the platform. Watch your step!
To solve this problem, we can use the equations of motion for an object in free fall.
(a) First, let's find the time it takes for the ball to hit the ground.
Given:
- Initial vertical position, h = 3 meters
- Vertical velocity, vy = 0 m/s (since the ball is rolling horizontally)
- Acceleration due to gravity, g = 10 m/s^2
We can use the equation: h = vy * t + 0.5 * g * t^2
Substituting the given values, we have: 3 = 0 * t + 0.5 * 10 * t^2
Simplifying the equation: 3 = 5t^2
Dividing both sides by 5: t^2 = 0.6
Taking the square root of both sides: t ≈ √0.6
Solving for t: t ≈ 0.775 seconds
Therefore, it takes approximately 0.775 seconds for the ball to hit the ground.
(b) Now, let's find the horizontal distance from the base of the platform where the ball hits the ground.
Given:
- Horizontal velocity, vx = 4.5 m/s
- Time, t = 0.775 seconds (calculated in part (a))
We can use the equation: d = vx * t
Substituting the given values, we have: d = 4.5 * 0.775
Simplifying the equation: d ≈ 3.4875 meters
Therefore, the ball hits the ground approximately 3.4875 meters from the base of the platform.
To find the time it takes for the ball to hit the ground, we can use the kinematic equation:
y = y0 + v0y * t + 1/2 * a * t^2
Where:
y is the vertical position of the ball (0 m, as it hits the ground)
y0 is the initial vertical position of the ball (3 m, the height of the platform)
v0y is the vertical component of the initial velocity of the ball (0 m/s, since the ball is not moving vertically at the start)
a is the acceleration due to gravity (-10 m/s², considering it is directed downwards)
t is the time we are looking for
By substituting the known values into the equation, we get:
0 = 3 + 0 * t + 1/2 * (-10) * t^2
Simplifying the equation gives:
0 = 3 - 5t^2
Rearranging the equation gives:
5t^2 = 3
t^2 = 3/5
Taking the square root of both sides, we find:
t ≈ √(3/5)
t ≈ 0.7746 seconds
Therefore, it takes approximately 0.7746 seconds for the ball to hit the ground.
Now, to find the horizontal distance the ball travels before hitting the ground, we can use the formula:
x = v0x * t
Where:
x is the horizontal distance
v0x is the horizontal component of the initial velocity of the ball (4.5 m/s, given in the question)
t is the time calculated earlier
Substituting the given values into the equation, we get:
x = 4.5 * 0.7746
Calculating this gives:
x ≈ 3.4746 meters
Therefore, the ball hits the ground approximately 3.4746 meters away from the base of the platform.
(a) 5t^2 = 3
(b) since the horizontal speed is constant, and distance = speed * time, that would be 4.5t (with t from (a) above)