two tall buildings are 200m apart .with what speed must a ball be thrown horizontally from the window 540 m above the ground in one building, so that it will enter a window 50m above the ground in the other?
Using Range = u^2*2h/g (using g = 9.8m/s^2)
u^2 = r*g/2h
u^2 = 400
u=20m/s
vertical problem
how long to fall 490 meters?
490 = (1/2)(9.81) t^2 solve for t
horizontal problem, same t
u t = 200
u = 200/t
Gfhf
To find the speed at which the ball must be thrown horizontally, we can use the principles of projectile motion and kinematics.
First, let's identify the given information:
- Distance between the two buildings (d) = 200m
- Height of the first building's window from the ground (h1) = 540m
- Height of the second building's window from the ground (h2) = 50m
Now, let's determine the time it takes for the ball to travel horizontally between the two buildings. Since there is no vertical acceleration horizontally, the time taken to travel horizontally is the same for both buildings.
The formula to calculate time (t) is:
t = d / v, where v is the horizontal velocity of the ball.
We can rearrange the formula to solve for v:
v = d / t (Equation 1)
Next, let's determine the time it takes for the ball to fall vertically from the first building to the second building.
The initial vertical velocity (u) is 0 since the ball is thrown horizontally, and the final vertical displacement (s) is the difference in height between the two windows:
s = h1 - h2 = 540m - 50m = 490m
The formula to calculate the time in free fall for an object with an initial vertical velocity of 0 is:
s = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values into the equation:
490 = (1/2) * 9.8 * t^2
Simplifying the equation:
t^2 = (490 * 2) / 9.8
t^2 = 100
t = √100 = 10s
Now, substitute the value of t into Equation 1 to find the horizontal velocity (v):
v = d / t
v = 200m / 10s
v = 20m/s
Therefore, the ball must be thrown horizontally with a speed of 20 m/s to enter the window on the other building.