A ball is thrown horizontally of a cliff. If the initial speed of the ball is 24.0 m/s and the cliff is 124.0 m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and rounded to three significant figures.
u = 24 m/s forever
x = u t where t is time to fall 124 m
h = (1/2)g t^2
124 = 4.9 t^2
t = 5.03 seconds
so
x = 24 * 5.03 = 121 meters
Well, that ball seems to have quite the high dive! Let's see where it lands, shall we?
Since the ball is thrown horizontally, we can ignore the initial velocity in the vertical direction. The only force acting on the ball is gravity, causing it to fall straight down.
Now, let's find the time it takes for the ball to hit the water. We can use the equation:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity, and t is the time.
Plugging in the values, we get:
124.0 = (1/2) * 9.8 * t^2
Simplifying that equation, we find:
t^2 = 124.0 * 2 / 9.8
t^2 = 25.183
Taking the square root of both sides, we get:
t ≈ 5.018
So, it takes approximately 5.018 seconds for the ball to hit the water.
Now, let's find out the distance from the base of the cliff where the ball will land. We know that distance equals velocity multiplied by time:
d = v * t
Plugging in the values, we get:
d = 24.0 * 5.018
d ≈ 120.432
Rounded to three significant figures, the ball will land about 120.432 meters from the base of the cliff.
Hope that answers your question in an amusing way!
To solve this problem, we can use the kinematic equation for vertical motion:
y = y0 + v0y * t + (1/2) * a * t^2
where:
- y is the final height (which we want to find)
- y0 is the initial height (124.0 m)
- v0y is the initial vertical velocity (0 m/s since the ball is thrown horizontally)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
Since the ball is thrown horizontally, the initial vertical velocity (v0y) is 0 m/s.
Hence, the equation becomes:
y = y0 + (1/2) * a * t^2
We know that the initial vertical position (y0) is 124.0 m and the acceleration due to gravity (a) is -9.8 m/s^2.
To find the time it takes for the ball to fall, we can use the equation:
y = y0 + v0 * t + (1/2) * a * t^2
Since the ball is thrown horizontally, the initial vertical velocity (v0y) is 0 m/s.
Thus, the equation becomes:
y = y0 + (1/2) * a * t^2
Plugging in the values:
124 = 0 + (1/2) * (-9.8) * t^2
Simplifying:
124 = -4.9 * t^2
Dividing both sides by -4.9:
t^2 = 124 / -4.9
t^2 = -25.30612245
Since time cannot be negative, we disregard the negative value.
t = sqrt(25.30612245)
t ≈ 5.031
Now, we can calculate the horizontal distance using the equation:
distance = velocity * time
Plugging in the values:
distance = 24.0 * 5.031
distance ≈ 120.744
Therefore, the ball will land approximately 120.744 meters from the base of the cliff. Rounded to three significant figures, the distance is 121.0 meters.
To solve this problem, we need to use the equations of motion in the horizontal and vertical directions for the ball.
In the horizontal direction, there is no acceleration acting on the ball, so its horizontal velocity remains constant throughout the motion. This means that the time taken to reach the water below is the same as the time taken to cover the horizontal distance.
The formula to calculate the time of flight is given by:
time = distance / initial horizontal velocity
Now, let's calculate the time of flight. Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s, and the acceleration due to gravity is acting vertically downwards.
In the vertical direction, we can use the equation of motion:
distance = (initial vertical velocity * time) + (0.5 * acceleration due to gravity * time^2)
Since the ball starts from rest in the vertical direction, the equation simplifies to:
distance = (0.5 * acceleration due to gravity * time^2)
We can rearrange this equation to solve for time:
time = sqrt((2 * distance) / acceleration due to gravity)
Given:
Initial vertical velocity (vy) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Distance (d) = 124.0 m
Let's use the formula to find the time of flight:
time = sqrt((2 * d) / g)
= sqrt((2 * 124.0) / 9.8)
= sqrt(24.98)
≈ 4.999
The time of flight is approximately 5 seconds.
Now, we can find the horizontal distance (x) using the formula:
x = initial horizontal velocity * time
= 24.0 * 5
= 120.0 m
Therefore, the ball will land approximately 120.0 meters from the base of the cliff in the water below.