A marble is thrown horizontally with a speed of 13.9 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 40 ° with the horizontal. From what height above the ground was the marble thrown?
Vertical problem:
How fast does it fall?
(1/2) m v^2 = m g h
so
v = sqrt (2 g h)
Horizontal problem:
u = constant = 13.9
so
tan 40 = sqrt(2 g h) /13.9
To find the height from which the marble was thrown, we need to analyze the motion of the marble. Let's break down the problem into different parts:
1. Find the time it takes for the marble to hit the ground:
The horizontal component of velocity remains constant throughout the motion, so we can use it to calculate the time.
Given:
Initial horizontal velocity (Vx) = 13.9 m/s
Launch angle (θ) = 40°
The horizontal component of velocity remains constant, so we can use this equation to find the time (t):
Vx = initial horizontal velocity * cos(angle)
Vx = 13.9 m/s * cos(40°)
Vx = 13.9 m/s * 0.766
Vx = 10.64 m/s
Now, we can determine the time it takes for the marble to hit the ground. We consider the vertical component of the motion:
h = initial vertical displacement
Vy = initial vertical velocity = 0 m/s (at the highest point)
g = acceleration due to gravity = 9.8 m/s²
The equation to calculate time is:
h = initial vertical displacement = Vy * t + (1/2) * g * t²
0 = 0 * t + (1/2) * 9.8 * t²
0 = 4.9 * t²
Since the marble hits the ground, the time will be positive.
Solving the equation:
4.9 * t² = 0
t² = 0
t = 0
From this calculation, we see that the marble hits the ground instantly. This might indicate that there is an error in the problem statement. Please double-check the problem and provide the correct information if necessary.