A baseball is hit with a speed of 27 meters per second at an angle of 45 degrees. It lands on the flat roof of a 13 meter tall building. If the ball was hit when it was 1 meter above the ground, what horizontal distance does it travel before it lands on the building?
A long jumper leaves the ground at 45 degrees above the horizontal and lands 8 meters away.
What is his "takeoff" speed?
A long jumper is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10 meters away horizontally, and 2.5 meters vertically below. If he jumps from the edge of the left bank at a 45 degree angle with the speed calculated in question.
How long or short from the opposite bank will he land?
#2: 8.0 m = (Vo^2/g)sin(2A) = Vo^2/g
Solve for Vo
For #1, Vx = Voy = 27/sqrt2 = 19.1 m/s
Y = 1 + Voy*t -(g/2)t^2 = 13
Solve for t. Then, use
X = Vx*t for the horizontal distance
For #3, use what you have learned and try it yourself. I assume you are supposed to use the takeoff speed calculated in the previous question.
For #2 What formula did you use when writing: (Vo^g)sin (2A)? Where did the 2 in the 2A come from?
For #1 Where did the 27/sqrt2 come from?
For #3 I still need help doing this one. I am not yet at the point where I can do it myself.
To find the horizontal distance that the baseball travels before landing on the building, we can use the principles of projectile motion.
1. First, let's calculate the time it takes for the baseball to reach the height of the building (13 meters) using the equation for vertical motion:
h = ut + (1/2)gt^2
where:
h = vertical displacement (13 meters)
u = initial vertical velocity (27 sin(45) m/s)
t = time
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation:
t = √(2h/g)
Substituting the values:
t = √(2 * 13 / 9.8)
t ≈ 1.47 seconds
2. Next, let's find the horizontal distance traveled by the baseball in this time:
d = ut
where:
d = horizontal distance
u = initial horizontal velocity (27 cos(45) m/s)
t = time
Substituting the values:
d = (27 cos(45)) * 1.47
d ≈ 27.84 meters
Therefore, the baseball travels approximately 27.84 meters horizontally before landing on the building.
Now let's move on to the second question:
To find the "takeoff" speed of the long jumper, we can use the range formula of projectile motion.
1. The range (distance) of the long jump can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
where:
R = range (8 meters)
v = takeoff speed (unknown)
θ = angle of takeoff (45 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation:
v = √((R * g) / sin(2θ))
Substituting the values:
v = √((8 * 9.8) / sin(90))
v ≈ 8.89 m/s
Therefore, the takeoff speed of the long jumper is approximately 8.89 m/s.
Lastly, let's answer the third question:
To determine how long or short the long jumper will land from the opposite bank, we need to calculate the horizontal distance traveled during the jump. We can use the range formula again.
1. Using the same formula as before:
R = (v^2 * sin(2θ)) / g
where:
R = range (unknown)
v = takeoff speed (8.89 m/s)
θ = angle of takeoff (45 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation:
R = (v^2 * sin(2θ)) / g
Substituting the values:
R = (8.89^2 * sin(90)) / 9.8
R ≈ 6.40 meters
Therefore, the long jumper will land approximately 6.40 meters short of the opposite bank.
To find the horizontal distance the baseball travels before it lands on the building, we can use the following steps:
Step 1: Break down the initial velocity of the baseball into its horizontal and vertical components.
- The initial velocity of the baseball is 27 meters per second at an angle of 45 degrees.
- The horizontal component of the velocity can be found using the formula Vx = V * cos(theta), where V is the magnitude of the velocity and theta is the launch angle.
- In this case, Vx = 27 * cos(45) = 19.091 meters per second.
Step 2: Calculate the time it takes for the baseball to reach the building.
- The vertical component of the velocity can be found using the formula Vy = V * sin(theta), where V is the magnitude of the velocity and theta is the launch angle.
- In this case, Vy = 27 * sin(45) = 19.091 meters per second.
- We can use the formula h = Vyt - 0.5 * g * t^2, where h is the vertical distance traveled, Vy is the initial vertical velocity, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time taken.
- In this case, h = 13 meters (height of the building), Vy = 19.091, and g = 9.8.
- Plugging in these values and solving the equation, we can find the time taken.
Step 3: Calculate the horizontal distance traveled.
- The horizontal distance traveled can be found using the formula d = Vx * t, where Vx is the horizontal component of the velocity and t is the time taken.
- In this case, Vx = 19.091 and t is the time calculated in Step 2.
- Plugging in these values and solving the equation, we can find the horizontal distance traveled.
For the second question, to find the takeoff speed of the long jumper, we can use the following steps:
Step 1: Break down the initial distance traveled by the long jumper into horizontal and vertical components.
- The initial horizontal distance traveled is given as 8 meters.
- The initial vertical distance traveled should be calculated based on the launch angle of 45 degrees.
Step 2: Calculate the horizontal and vertical components of the velocity.
- We can use the formula dx = V * cos(theta) * t, where dx is the horizontal distance, V is the magnitude of the velocity, theta is the launch angle, and t is the time taken.
- In this case, dx = 8 meters, theta = 45 degrees, and we need to solve for V.
- Rearranging the formula, V = dx / (cos(theta) * t).
For the third question, to find how long or short the long jumper will land from the opposite bank, we can use the following steps:
Step 1: Calculate the horizontal distance traveled by the long jumper.
- The horizontal distance traveled can be calculated using the formula dx = V * cos(theta) * t, where dx is the horizontal distance, V is the magnitude of the velocity, theta is the launch angle, and t is the time taken.
- In this case, dx is the distance between the left bank and the right bank, theta = 45 degrees, and V is the takeoff speed calculated in the previous question.
- Plugging in these values and solving the equation, we can find the horizontal distance traveled.
Step 2: Calculate the vertical distance traveled by the long jumper.
- The vertical distance traveled can be calculated using the formula dy = V * sin(theta) * t + 0.5 * g * t^2, where dy is the vertical distance, V is the magnitude of the velocity, theta is the launch angle, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time taken.
- In this case, dy is the difference in vertical height between the left and right banks, theta = 45 degrees, V is the takeoff speed calculated in the previous question, and we need to solve for t.
- Rearranging the formula, t = (dy - 0.5 * g * t^2) / (V * sin(theta)).
Step 3: Calculate the horizontal displacement from the opposite bank.
- The horizontal displacement from the opposite bank can be found using the previously calculated horizontal distance traveled and the horizontal distance between the left and right banks.
- In this case, the horizontal displacement is the difference between the horizontal distance traveled and the distance between the left and right banks.