5)A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the hill would the ball hit the ground?
45.0 m
6)A missile launched at a velocity of 30.0 m/s at an angle of 30.0 to the normal. What is the maximum height the missile attains?
11.5 m
5)
I did not get the same answer you got.
Since there is no vertical component, the time to hit the ground is the same as if th ball had been dropped from a height of 50.0m. Calculate how long it takes for the ball to fall 50.0m. Take that time and multiply by the horizontal velocity (given as 10.0 m/s).
6)
I did not get the same answer you had.
Again, break the problem into vertical and horizontal components. Find the initial vertical component (30.0 m/s) * sin(60) because the problem says the missile is launched 30 degress from the NORMAL.
Determine how long before the vertical velocity is 0. That will be the maximum height. Plug that time into the distance equation:
t = time
g=gravity acceleration
distance = (initial velocity) * t + (1/2)(g)(t^2)
Remember that for positive up, gravity will be a negative acceleration.
So, pay attention to the sign of the second term.
To answer question 5:
Step 1: Determine the time it takes for the ball to hit the ground.
Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. The only force acting on the ball in the vertical direction is gravity, which causes it to accelerate downward at a rate of 9.8 m/s^2.
We can use the formula: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity, and t is the time taken.
Given that the initial height is 50.0 m and the acceleration due to gravity is 9.8 m/s^2, we can solve for t:
50.0 = (1/2) * 9.8 * t^2
t^2 = 10.2
t = sqrt(10.2) ≈ 3.19 s
Step 2: Calculate the horizontal distance traveled by the ball.
The horizontal distance is given by the formula: d = v * t, where d is the distance, v is the horizontal velocity, and t is the time.
Given that the horizontal velocity is 10.0 m/s and the time is 3.19 s, we can calculate the horizontal distance:
d = 10.0 * 3.19 ≈ 31.9 m
Therefore, the ball would hit the ground approximately 31.9 meters from the base of the hill.
To answer question 6:
Step 1: Resolve the missile's initial velocity into its horizontal and vertical components.
The horizontal component of the velocity (Vx) is given by: Vx = v * cos(theta), where v is the magnitude of the velocity and theta is the launch angle.
Given that the magnitude of the velocity is 30.0 m/s and the launch angle is 30.0 degrees, we can calculate the horizontal component:
Vx = 30.0 * cos(30.0) ≈ 25.98 m/s
The vertical component of the velocity (Vy) is given by: Vy = v * sin(theta), where v is the magnitude of the velocity and theta is the launch angle.
Given that the magnitude of the velocity is 30.0 m/s and the launch angle is 30.0 degrees, we can calculate the vertical component:
Vy = 30.0 * sin(30.0) ≈ 15.0 m/s
Step 2: Determine the time it takes for the missile to reach its maximum height.
The time to reach maximum height (t_max) can be found using the formula: t_max = Vy / g, where Vy is the vertical component of the velocity and g is the acceleration due to gravity.
Given that the vertical component is 15.0 m/s and the acceleration due to gravity is 9.8 m/s^2, we can calculate the time to reach maximum height:
t_max = 15.0 / 9.8 ≈ 1.53 s
Step 3: Calculate the maximum height reached by the missile.
The maximum height (h_max) can be found using the formula: h_max = Vy * t_max - (1/2) * g * t_max^2, where Vy is the vertical component of the velocity, t_max is the time to reach maximum height, and g is the acceleration due to gravity.
Given that the vertical component is 15.0 m/s, the time to reach maximum height is 1.53 s, and the acceleration due to gravity is 9.8 m/s^2, we can calculate the maximum height:
h_max = 15.0 * 1.53 - (1/2) * 9.8 * 1.53^2 ≈ 11.5 m
Therefore, the missile attains a maximum height of approximately 11.5 meters.
To solve question 5, we can use kinematic equations. The projectile is thrown horizontally, so the initial vertical velocity is 0 m/s. The only force acting on the ball is gravity, which causes it to accelerate vertically downwards at a rate of 9.8 m/s².
We can use the kinematic equation for vertical displacement:
y = y₀ + v₀yt + (1/2)gt²
where
y = vertical displacement (final position - initial position)
y₀ = initial vertical position (50.0 m)
v₀y = vertical component of initial velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s²)
t = time (unknown)
Plugging in the values, the equation becomes:
0 = 50.0 + 0 × t + (1/2)(-9.8)t²
Simplifying the equation, we get:
0 = 50.0 - 4.9t²
Rearranging the equation, we get:
t² = 50.0 / 4.9
t² ≈ 10.2
Taking the square root of both sides, we get:
t ≈ √10.2
t ≈ 3.2 s
The time it takes for the ball to hit the ground is approximately 3.2 seconds. Since the ball is thrown horizontally at a constant speed of 10.0 m/s, the horizontal distance traveled can be calculated by multiplying the time by the horizontal velocity:
distance = velocity × time
distance = 10.0 × 3.2
distance ≈ 32 m
Therefore, the ball would hit the ground approximately 32 meters from the base of the hill.
To solve question 6, we can use the projectile motion equations. The missile is launched at an angle of 30.0° to the normal, and its initial velocity is 30.0 m/s. We need to find the maximum height the missile attains.
The vertical motion of the missile can be modeled using the equation:
y = y₀ + v₀y × t + (1/2)gt²
where
y = vertical displacement (unknown)
y₀ = initial vertical position (0 m)
v₀y = vertical component of initial velocity (30.0 m/s × sin 30.0°)
g = acceleration due to gravity (-9.8 m/s²)
t = time (unknown)
Plugging in the values, the equation becomes:
y = 0 + 30.0 × sin 30.0° × t + (1/2)(-9.8)t²
Since we're trying to find the maximum height, we can determine the time it takes to reach the highest point of the trajectory using:
vfy = v₀y + gt
where
vfy = vertical component of final velocity (0 m/s at the highest point)
v₀y = vertical component of initial velocity (30.0 m/s × sin 30.0°)
g = acceleration due to gravity (-9.8 m/s²)
t = time (unknown)
Plugging in the values, the equation becomes:
0 = 30.0 × sin 30.0° + (-9.8)t
Simplifying the equation, we get:
t = -30.0 × sin 30.0° / -9.8
t ≈ 1.5 s
Substituting the value of t back into the y equation:
y = 0 + 30.0 × sin 30.0° × 1.5 + (1/2)(-9.8)(1.5)²
Simplifying the equation, we get:
y ≈ 11.5 m
Therefore, the maximum height the missile attains is approximately 11.5 meters.