A)A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.15 m/s at an angle of 19.0o below the horizontal. It strikes the ground 2.50 s later. How far horizontally from the base of the building does the ball strike the ground?
B)Calculate the height from which the ball was thrown.
C)How long does it take the ball to reach a point 12.0 m below the level of launching?
Vo = 8.15m/s[-19o]
Xo = 8.15*cos(-19) = 7.71 m/s.
Yo = 8.15*sin(-19) = -2.65 m/s.
A.Dx=Xo * Tf = 7.71m/s * 2.50s=19.26 m/s
B. h = Yo*T + 0.5g*T^2
h = -2.65*2.5 + 4.9*2.5^2 = 24 m.
C. -2.65*T + 4.9T^2 = 24-12 = 12
4.9T^2 - 2.65T - 12 = 0
T = 1.86 s.
To solve these problems, we can use the equations of motion for projectile motion.
A) To find the horizontal distance the ball travels, we need to find the horizontal component of its velocity. We can use the equation:
Vx = V0 * cos(θ)
Where Vx is the horizontal component of velocity, V0 is the initial velocity of the ball, and θ is the angle at which it is thrown.
Given:
V0 = 8.15 m/s
θ = 19.0°
Plugging in these values, we can find Vx:
Vx = 8.15 * cos(19.0)
= 7.746 m/s
Next, we can use the equation for horizontal distance traveled by a projectile:
Horizontal distance = Vx * time
Given:
time = 2.50 s
Plugging in the values, we have:
Horizontal distance = 7.746 * 2.50
= 19.364 m
So, the ball strikes the ground approximately 19.364 meters horizontally from the base of the building.
B) To calculate the height from which the ball was thrown, we can use the equation for vertical distance traveled by a projectile:
Vertical distance = Vy * time + (0.5 * g * time^2)
Where Vy is the vertical component of velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and time is the time it takes for the ball to reach the ground.
Given:
time = 2.50 s
First, let's find Vy using:
Vy = V0 * sin(θ)
Vy = 8.15 * sin(19.0)
= 2.865 m/s
Plugging in the values, we have:
Vertical distance = 2.865 * 2.5 + (0.5 * 9.8 * 2.5^2)
= 7.163 m
So, the height from which the ball was thrown is approximately 7.163 meters.
C) To find how long it takes for the ball to reach a point 12.0 m below the level of launching, we need to solve the equation:
Vertical distance = Vy * time + (0.5 * g * time^2)
Given:
Vertical distance = -12.0 m (negative because it is below the level of launching)
We know Vy is the same as before:
Vy = 2.865 m/s
Plugging in the values, we have:
-12.0 = 2.865 * time + (0.5 * 9.8 * time^2)
This is a quadratic equation that can be solved to find the time it takes for the ball to reach 12.0 m below the level of launching. Solving this equation will give us two possible values of time, but since time cannot be negative, we can discard the negative value.
After solving the equation, we find:
time ≈ 1.5 s
So, it takes approximately 1.5 seconds for the ball to reach a point 12.0 m below the level of launching.