A baseball is hit with a speed of 27.0 m/s at an angle of 41.0 ∘ . It lands on the flat roof of a 14.0 m -tall nearby building. If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
Horizontal problem:
u = horizontal speed = 27 cos 42 forever
so horizontal distance is u T
to find T do vertical problem
vertical problem
Vi = 27 sin 42
Hi = 1.3
h = 14 at t = T
h = Hi + Vi t - 4.9 t^2
so at h = 14 and t = T
14 = 1.3 + 27 sin 42 T - 4.9 T^2
solve quadratic for T
distance = u T
To find the horizontal distance the baseball travels before it lands on the building, we can use the equations of motion to analyze the projectile motion.
First, let's break down the initial velocity of the baseball into its horizontal and vertical components.
The vertical component of the initial velocity can be found using the equation:
Viy = Vi * sin(theta)
where Vi is the initial speed and theta is the launch angle.
Viy = 27.0 m/s * sin(41.0°)
Viy ≈ 17.8 m/s
The horizontal component of the initial velocity is given by:
Vix = Vi * cos(theta)
Vix = 27.0 m/s * cos(41.0°)
Vix ≈ 21.0 m/s
Now, let's determine the time it takes for the baseball to reach the height of the building. We can use the equation:
hf = hi + Viy*t + (1/2)*g*t^2
where hf is the final height (14.0 m), hi is the initial height (1.3 m), Viy is the vertical component of the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Rearranging the equation to solve for time (t):
hf - hi = Viy*t + (1/2)*g*t^2
(14.0 m - 1.3 m) = 17.8 m/s * t + (1/2)(-9.8 m/s^2) * t^2
12.7 m ≈ 17.8 m/s * t - 4.9 m/s^2 * t^2
4.9 m/s^2 * t^2 - 17.8 m/s * t + 12.7 m ≈ 0
Solving this quadratic equation will give us the time of flight. We can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 4.9 m/s^2, b = -17.8 m/s, and c = 12.7 m.
Calculating the discriminant (b^2 - 4ac):
discriminant = (-17.8 m/s)^2 - 4 * 4.9 m/s^2 * 12.7 m
discriminant ≈ 127.14 m^2/s^2
Since the discriminant is positive, there are two real solutions for t.
Using the quadratic formula:
t = (-(-17.8 m/s) ± sqrt(127.14 m^2/s^2)) / (2 * 4.9 m/s^2)
t1 = (17.8 m/s + sqrt(127.14 m^2/s^2)) / (2 * 4.9 m/s^2)
t1 ≈ 3.9 s
t2 = (17.8 m/s - sqrt(127.14 m^2/s^2)) / (2 * 4.9 m/s^2)
t2 ≈ 0.7 s
Since the time of flight cannot be negative, we discard t2 and only use t1 ≈ 3.9 s.
Finally, we can find the horizontal distance using the equation:
d = Vix * t
where d is the horizontal distance and Vix is the horizontal component of the initial velocity.
d = 21.0 m/s * 3.9 s
d ≈ 81.9 m
Therefore, the horizontal distance the baseball travels before it lands on the building is approximately 81.9 meters.
To find the horizontal distance that the baseball travels before it lands on the building, we need to break down the problem into its horizontal and vertical components.
First, let's consider the vertical motion of the ball. The ball is initially 1.3 m above the ground and lands on a 14.0 m tall building. The maximum height reached by the ball is the difference between the height of the building and its initial height: 14.0 m - 1.3 m = 12.7 m.
Now, let's calculate the time it takes for the ball to reach its maximum height. We can use the vertical motion formula:
vf = vi + gt
where:
vf = final vertical velocity (0 m/s, at the highest point)
vi = initial vertical velocity (27.0 m/s * sin(41°))
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the formula, we have:
t = (vf - vi) / g
Substituting the values, we get:
t = (0 m/s - 27.0 m/s * sin(41°)) / -9.8 m/s^2
Using a calculator, we find that t ≈ 2.26 s.
Next, let's calculate the time it takes for the ball to fall back to the ground from its maximum height. We can use the same vertical motion formula, but this time with the initial velocity being 0 m/s:
t = (vf - vi) / g
Substituting the values, we have:
t = (-27.0 m/s * sin(41°) - 0 m/s) / -9.8 m/s^2
Again, using a calculator, we find that t ≈ 2.26 s.
Since the time for the ball to reach its maximum height and the time for it to fall back down are equal, the total time of flight is t + t = 2.26 s + 2.26 s = 4.52 s.
Finally, we can find the horizontal distance traveled by the ball using the horizontal motion formula:
d = v * t
where:
d = horizontal distance
v = horizontal velocity (27.0 m/s * cos(41°))
t = time of flight (4.52 s)
Substituting the values, we have:
d = (27.0 m/s * cos(41°)) * 4.52 s
Using a calculator, we find that d ≈ 101.3 m.
Therefore, the horizontal distance the baseball travels before it lands on the building is approximately 101.3 meters.