A marble is thrown horizontally with a speed of 10.5 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 34.1 ° with the horizontal. From what height above the ground was the marble thrown?
tan φ=v(y)/v(x) ,
v(y)=v(x)• tanφ.
v(y)=g•t, => t=v(y)/g
h=gt²/2 =g• v²(y)/2•g²
= v²(y)/2•g= (v(x)• tanφ)²/ 2•g= (10.5•tan 34.1º)²/2•9.8 =2.6 m.
The answer is correct, but the given data are strange enough , especially, the angle…
To determine the height from which the marble was thrown, we can use the equations of motion. In this case, we'll use the vertical motion equation.
1. First, let's find the time it takes for the marble to reach the ground. Since the initial velocity in the vertical direction is 0 (it's thrown horizontally), we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Rearranging the equation for the initial velocity, we get:
u = -(v - at)
Substituting the known values:
0 = -(0 - 9.8t)
0 = 9.8t
Solving for t, we find t = 0 seconds.
This means that it takes 0 seconds for the marble to reach the ground.
2. Now, we know the time it takes for the marble to reach the ground. We can use this to find the height from which it was thrown using the equation:
s = ut + (1/2)at^2
Where:
s = height (unknown)
u = initial velocity in the vertical direction (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (0 seconds)
Plugging in the values, we get:
s = ut + (1/2)at^2
s = 0t + (1/2)(-9.8)(0)^2
s = 0
Therefore, the marble must have been thrown from a height of 0 meters above the ground.
To solve this problem, we can use the equations of motion to find the height at which the marble was thrown.
Let's consider the horizontal and vertical components of the marble's motion separately.
In the horizontal direction, the marble is thrown horizontally and continues to move horizontally until it reaches the ground. Therefore, the horizontal component of velocity remains constant throughout and is given as:
Vx = 10.5 m/s
In the vertical direction, the marble is subject to gravitational acceleration. We can use the following kinematic equation to find the initial vertical velocity (Vy) when the marble was thrown:
Vy^2 = Vyo^2 + 2gΔy
Since the marble is thrown horizontally, the initial vertical velocity (Vyo) is zero (since there is no initial vertical motion). Also, the final vertical velocity (Vy) when the marble strikes the ground can be calculated using the angle (θ) made with the horizontal:
Vy = V * sin(θ)
Therefore, we have:
Vy^2 = (V * sin(θ))^2 = Vx^2 + 2gΔy
Using the given values, we can substitute the known variables:
(V * sin(θ))^2 = Vx^2 + 2gΔy
(10.5 * sin(34.1))^2 = (10.5)^2 + 2 * 9.8 * Δy
Now, we can solve this equation for Δy, which represents the height at which the marble was thrown from.
First, calculate (V * sin(θ))^2:
(10.5 * sin(34.1))^2 ≈ 10.5^2 + 2 * 9.8 * Δy
(5.687)^2 ≈ 110.25 + 19.6Δy
32.323769 ≈ 110.25 + 19.6Δy
Next, rearrange the equation to isolate Δy:
19.6Δy ≈ 110.25 - 32.323769
19.6Δy ≈ 77.926231
Δy ≈ 77.926231 / 19.6
Δy ≈ 3.976
Therefore, the marble was thrown from a height of approximately 3.976 meters above the ground.