A playful college student tosses a water balloon out of his dorm window at a speed of 6.2 m/s at an angle of 10° above the horizontal. The dorm window is 11 m above ground level.
1)What is the maximum height above the ground of the balloon?
2) At what time does the balloon strike the ground?
3) At what distance from the dorm does the balloon hit the ground?
4) With what angle does the balloon strike the ground?
the answer is wrongg ;(
To answer these questions, we can analyze the motion of the water balloon using the equations of projectile motion. We'll begin by breaking the initial velocity into its horizontal and vertical components.
Given:
- Initial speed of the balloon (v₀) = 6.2 m/s
- Launch angle (θ) = 10°
- Height of the dorm window (h) = 11 m
- Acceleration due to gravity (g) = 9.8 m/s²
1) To find the maximum height above the ground of the balloon, we can use the vertical component of its motion. The maximum height (H) can be found using the equation:
H = (v₀*sin(θ))² / (2*g)
Substituting in the given values:
H = (6.2*sin(10°))² / (2*9.8)
H ≈ 1.04 m
Therefore, the maximum height above the ground of the balloon is approximately 1.04 m.
2) To determine the time it takes for the balloon to strike the ground, we can use the equation of vertical motion:
h = v₀y*t - (1/2)*g*t²
Since the final height is 0, we solve the equation for t:
0 = (v₀*sin(θ)) * t - (1/2)*g*t²
Rearranging the equation:
(1/2)*g*t² = (v₀*sin(θ)) * t
Simplifying:
(1/2)*9.8*t² = 6.2*sin(10°)*t
Dividing both sides by t:
(1/2)*9.8*t = 6.2*sin(10°)
Solving for t:
t = (6.2*sin(10°)) / ((1/2)*9.8)
t ≈ 1.06 s
Therefore, the balloon strikes the ground at approximately 1.06 seconds after it was launched.
3) To calculate the horizontal distance travelled by the balloon, we can use the equation of horizontal motion:
d = v₀x * t
The horizontal speed of the balloon (v₀x) can be found using:
v₀x = v₀ * cos(θ)
Using the given values:
v₀x = 6.2 * cos(10°)
v₀x ≈ 6.04 m/s
Plugging these values into the equation of horizontal motion:
d = 6.04 * 1.06
d ≈ 6.40 m
Therefore, the balloon hits the ground at a distance of approximately 6.40 meters from the dorm.
4) Finally, to determine the angle at which the balloon strikes the ground, we can use trigonometry. The angle (α) can be found using:
tan(α) = (v₀y - gt) / v₀x
Plugging in the given values:
tan(α) = (6.2*sin(10°) - 9.8*1.06) / 6.2*cos(10°)
Simplifying:
tan(α) ≈ -0.042
α ≈ -2.41°
Therefore, the balloon strikes the ground at an angle of approximately -2.41° relative to the horizontal.
To find the answers to these questions, we can use the equations of motion for projectile motion. The motion of the water balloon can be divided into horizontal and vertical components. We can use the following equations:
1) To find the maximum height above the ground of the balloon, we need to determine the vertical component of the initial velocity. We can use the equation:
Vy = V * sin(θ)
where:
- Vy is the vertical component of the initial velocity
- V is the magnitude of the initial velocity (6.2 m/s in this case)
- θ is the angle of the initial velocity (10° in this case)
So, Vy = 6.2 m/s * sin(10°)
Now, to find the maximum height, we need to use the kinematic equation for vertical displacement:
Δy = Vy^2 / (2 * g)
where:
- Δy is the vertical displacement (maximum height)
- Vy is the vertical component of the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
2) To find the time it takes for the balloon to strike the ground, we need to determine the time it takes for the vertical displacement to be equal to the initial height (11 m). We can use the kinematic equation for vertical displacement:
Δy = Vy * t - (1/2) * g * t^2
where:
- Δy is the vertical displacement (equal to the initial height, 11 m)
- Vy is the vertical component of the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time it takes for the balloon to strike the ground
3) To find the horizontal distance from the dorm where the balloon hits the ground, we can use the horizontal component of the initial velocity:
Vx = V * cos(θ)
where:
- Vx is the horizontal component of the initial velocity
- V is the magnitude of the initial velocity (6.2 m/s in this case)
- θ is the angle of the initial velocity (10° in this case)
Now, to find the distance, we can use the equation:
distance = Vx * t
where:
- distance is the horizontal distance from the dorm
- Vx is the horizontal component of the initial velocity
- t is the time it takes for the balloon to strike the ground
4) To find the angle at which the balloon strikes the ground, we can use the inverse tangent function:
θ' = tan^(-1)(Vy / Vx )
where:
- θ' is the angle at which the balloon strikes the ground
- Vy is the vertical component of the initial velocity
- Vx is the horizontal component of the initial velocity
Now, we can calculate the answers to the given questions using the above equations and the given values.
Vo = 6.2m/s[10o]
Xo = 6.2*cos10 = 6.106 m/s.
Yo = 6.2*sin10 = 1.077 m/s.
1. h = ho + (Y^2-Yo^2)/2g
h = 11 + (0-(1.077)^2)/-19.6 = 11.06 m.
Above gnd.
2. h = 0.5g*t^2 = 11.06 m.
4.9t^2 = 11.06
t^2 = 2.257
t = 1.50 s.
3. d = Xo * t = 6.106m/s * 1.50s=9.17 m.
4. Y^2=Yo^2 + 2g*h= 0 + 19.6*11.06=216.8
Y = 14.72 m/s.
Tan A = Y/Xo = 14.72/6.106 = 2.41129
A = 67.5o