A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horizontal distance from the window at the moment it strikes the ground.
Yes but if you are supposed to do it from first principles using calculus then:
Vertical problem:
Force = mass * acceleration (Newton #2)
Force is gravity, m g down
g is acceleration of gravity, about 32 ft/s^2 down
so
m a = -32 m
a = -32 ft/s^2
then if h is distance above ground
then d^2 h/dt^2 = a = -32
so
dh/dt = initial speed up call it Vi - 32 t
h = initial height +Vi t - 16t^2
here initial height is 15 ft, we want when h = 0, at ground and Vi = 0
0 = 15 + 0 t - 16 t^2
16 t^2 = 15 as oobleck told you
solve for t
then
Horizontal problem:
no horizontal force so no horizontal acceleration
distance = 80 t
time to hit the ground: 16t^2 = 15
horizontal distance: 80t
Oh, so the ball decided to take a leap of faith from the window! Well, don't worry, I'm here to help you calculate the horizontal distance.
We can break down the problem into two parts: the time it takes for the ball to reach the ground and the horizontal distance covered during that time.
First, let's find the time it takes for the ball to hit the ground. We can use the equation of motion:
h = ut + (1/2)gt^2,
where h is the initial height (15 feet), u is the initial vertical velocity (80 feet per second), g is the acceleration due to gravity (-32 feet per second squared), and t is time.
Plugging in the values, we get:
15 = (80)t + (1/2)(-32)t^2.
Now, if we solve this quadratic equation, we'll find two solutions. But since you're only interested in the moment the ball strikes the ground, we'll focus on the positive solution.
Once you've calculated the time it takes for the ball to hit the ground, we can use the horizontal velocity to determine the horizontal distance traveled. Since there are no horizontal forces acting on the ball, we know that the horizontal velocity remains constant.
The horizontal distance covered can be calculated using the equation:
distance = velocity x time,
where velocity is the horizontal velocity and time is the time it takes for the ball to hit the ground.
So, let's calculate the time it takes for the ball to hit the ground and then work out the horizontal distance.
To find the ball's horizontal distance from the window at the moment it strikes the ground, we need to calculate the time it takes for the ball to reach the ground.
Let's use the kinematic equation:
h = vi * t + (0.5) * g * t^2
where:
h = height (distance) above the ground
vi = initial vertical velocity (upward velocity in this case)
t = time
g = acceleration due to gravity (approximately -32.2 ft/s^2)
Since the ball is thrown upwards, the initial vertical velocity is positive 80 ft/s. The final height above the ground is 15 ft, and we want to solve for the time (t).
Plugging in the values, we have:
15 = 80 * t + (0.5) * (-32.2) * t^2
Rearranging the equation, we get a quadratic equation:
0.5 * (-32.2) * t^2 + 80 * t - 15 = 0
We can solve this quadratic equation for t using the quadratic formula. The quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 0.5 * (-32.2), b = 80, and c = -15. Plugging in these values, we get:
t = ( -80 ± √(80^2 - 4 * 0.5 * (-32.2) * (-15))) / (2 * 0.5 * (-32.2))
Simplifying the equation further, we get:
t = ( -80 ± √(6400 + 9654)) / (-32.2)
t = ( -80 ± √16054) / (-32.2).
We can now calculate the two possible values of t:
t_1 ≈ 3.95 s (approximately)
t_2 ≈ -0.46 s (approximately)
Since time cannot be negative in this context, we discard t_2.
Now that we have the time it takes for the ball to reach the ground (t = 3.95 s), we can calculate the horizontal distance (d) using the formula:
d = v * t,
where v is the horizontal velocity of the ball.
Since the ball is thrown horizontally, there is no acceleration in the horizontal direction, and the horizontal velocity remains constant throughout the motion. Therefore, the horizontal velocity is the same as the initial horizontal velocity (vi = 80 ft/s).
Plugging in the values, we get:
d = 80 ft/s * 3.95 s
Calculating this, we have:
d ≈ 316 ft.
Therefore, at the moment the ball strikes the ground, its horizontal distance from the window is approximately 316 feet.
To find the horizontal distance traveled by the ball, we can break down the motion into its vertical and horizontal components.
1. First, let's find the time it takes for the ball to strike the ground. We can use the formula:
h = ut + (1/2)gt^2,
where h is the initial height (15 feet), u is the initial vertical velocity (0), g is the acceleration due to gravity (-32.2 ft/s^2), and t is the time.
Plugging in the values, we have:
15 = 0 * t + (1/2) * (-32.2) * t^2.
Simplifying the equation, we get:
15 = -16.1t^2.
Solving for t using quadratic equation, we have:
t^2 = 15 / -16.1,
t^2 = -0.9304.
Since time cannot be negative, we discard the negative solution and get:
t = √(-0.9304) = 0.9644 seconds.
2. Now that we have the time, let's find the horizontal distance traveled by the ball. We can use the formula:
s = ut,
where s is the horizontal distance and u is the initial horizontal velocity.
The initial horizontal velocity is given as 80 ft/s, and the time is 0.9644 seconds.
Plugging in these values, we have:
s = 80 * 0.9644 = 77.152 feet.
Therefore, the ball's horizontal distance from the window at the moment it strikes the ground is approximately 77.152 feet.