A boy throws a ball off the top of a 30m high building with a speed of 10m/s at an angle of 30 degrees to the horizontal. How far from the base of the building does the ball strike?
I broke in up into 2 dimensional:
X horizontal values: Xi=0 Xf=? Vi=8.6m/s. Vf=? a=0
Y vertical values: Xi=0, Xf=15m, Vi=5, Vf? a=9.82. I then solved for time in the Y then used the distant formula to find my answer. 11.3 m away. Is this right?
vertical:
hf=hi+vi*t-gt^2 /3
0=30+10sin30*t-4.9t^2
time
4.9t^2-5t-30=0 I get about 3 sec
horizontal:
distance=10cos30 * 3.03 sec which is not your answer.
To solve this problem, you have properly broken it down into two dimensions, horizontal (x) and vertical (y), which is a good approach.
For the x-direction, since there is no acceleration (a = 0), the horizontal velocity remains constant throughout the motion. Given that the initial horizontal velocity (Vi) is 10 m/s, and the angle (θ) is 30 degrees, you need to calculate the horizontal component of the velocity (Vx) using trigonometry.
Vx = Vi * cos(θ)
= 10 m/s * cos(30 degrees)
= 10 m/s * 0.866
≈ 8.66 m/s
Therefore, the horizontal component of the velocity (Vx) is approximately 8.66 m/s.
Next, for the y-direction, it is important to consider that the ball is being launched vertically upward from a height of 30 m. Therefore, you need to determine the time it takes for the ball to reach the top of its trajectory and then fall back down to the ground.
To find the time of flight, you can use the equation of motion:
Δy = Viy * t - (1/2) * g * t^2
where Δy is the change in height, Viy is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.82 m/s^2), and t is the time of flight.
Since the initial vertical velocity (Viy) is given as 10 m/s * sin(30 degrees), and the change in height (Δy) is -30 m (negative since the ball is falling back down), you can rearrange the equation to solve for time (t):
-30 m = (10 m/s * sin(30 degrees)) * t - (1/2) * 9.82 m/s^2 * t^2
Simplifying the equation, you get:
-30 = 5 * t - 4.91 * t^2
Rearrange the equation to form a quadratic equation:
4.91 * t^2 - 5 * t - 30 = 0
You can solve this quadratic equation using the quadratic formula, or factoring, to find the positive value of t. The estimated value of t is approximately 4.14 s.
Now that you know the time of flight, you can calculate the horizontal distance traveled by the ball using the equation:
Δx = Vx * t
Substituting the values, you have:
Δx = 8.66 m/s * 4.14 s
≈ 35.842 m
Therefore, the ball strikes the ground approximately 35.842 meters away from the base of the building.