A projectile is launched with an initial speed of 59.0 m/s at an angle of 32.0° above the horizontal. The projectile lands on a hillside 3.80 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(a) What is the projectile's velocity at the highest point of its trajectory?
magnitude ? m/s

direction ? ° counterclockwise from the +x-axis

(b) What is the straight-line distance from where the projectile was launched to where it hits its target? in m

in the vertical direction

... h = 1/2 g t^2 + 59 sin(32º) t

in the horizontal direction
... d = 59 cos(32º) t

use the axis of symmetry of the parabola to find the time for max height

plug the time into the vertical equation to find the max height

the initial energy minus the gravitational potential at max height gives the energy (velocity) at max height

the projectile has no vertical velocity at max height, so the direction is 0º

plug the flight time into the vertical equation to find the impact height

plug the flight time into the horizontal equation to find the horizontal distance

use Pythagoras to find the straight line distance

(a) Oh boy, we've got a projectile on a hillside! Alright, to find the velocity at the highest point, we can use some simple equations. The horizontal component of the velocity remains constant throughout the entire trajectory, so it's just 59.0 m/s.

Now, for the vertical component, we can use some trigonometry. Since the angle is 32.0° above the horizontal, the vertical component is given by 59.0 m/s * sin(32.0°) = 31.5 m/s.

So, at the highest point of its trajectory, the projectile's velocity will be 31.5 m/s in the upward direction.

(b) Alright, let's calculate the distance it travels horizontally. The horizontal component of the velocity is 59.0 m/s, and the time it takes to land is 3.80 s. So, the distance traveled in the x-direction is 59.0 m/s * 3.80 s = 224.2 m.

Therefore, the straight-line distance from where the projectile was launched to where it hits its target is approximately 224.2 meters.

To solve part (a) of the problem, we need to find the projectile's velocity at the highest point of its trajectory. We can use the following equations of motion:

1. Vfy = Viy + g * t
2. Vfy^2 = Viy^2 + 2 * g * Δy

Where:
- Vfy and Viy are the final and initial velocities in the y-direction.
- g is the acceleration due to gravity (approximately 9.8 m/s²).
- t is the time of flight.
- Δy is the maximum height reached by the projectile.

At the highest point, the vertical velocity becomes zero, so we can set Vfy = 0.
Also, the time of flight can be divided in half to find the time when the projectile reaches its highest point (t/2).

Using equation 1, we can solve for Viy:
0 = Viy + (9.8 m/s²) * (t/2)

Simplifying, we get:
Viy = -(4.9 m/s²) * t

Using this value of Viy, we can substitute it into equation 2:
0 = (-(4.9 m/s²) * t)^2 + 2 * (9.8 m/s²) * Δy

Simplifying again, we get:
Δy = (Viy^2) / (2 * g)

Plugging in the values, we have:
Δy = (-(4.9 m/s²) * t)^2 / (2 * 9.8 m/s²)

Simplifying further, we get:
Δy = (-(4.9 m/s²) * (3.80 s))^2 / (2 * 9.8 m/s²)

Solving for Δy:
Δy ≈ -47.04 m

This means that the projectile reaches a maximum height of approximately 47.04 meters above the launch point.

To find the magnitude of the projectile's velocity at the highest point, we can use the following equation:

3. Vf = √(Vfx^2 + Vfy^2)

Since there is no acceleration in the horizontal direction, the horizontal component of the velocity remains constant throughout the flight.

Vfx = Vix = (59.0 m/s) * cos(32.0°)

Substituting this value into equation 3, the magnitude of the velocity at the highest point is:

Vf = √(Vfx^2 + Vfy^2)
= √((59.0 m/s * cos(32.0°))^2 + (-(4.9 m/s²) * (3.80 s))^2)

Calculating this, we find:
Vf ≈ 55.9 m/s

To determine the direction of the velocity, we need to find the angle counterclockwise from the +x-axis. We can use the equation:

θ = tan^(-1)(Vfy / Vfx)

Using the values we've already calculated, we have:

θ = tan^(-1)((-(4.9 m/s²) * (3.80 s)) / (59.0 m/s * cos(32.0°)))

Calculating this, we find:
θ ≈ -38.5°

Therefore, the projectile's velocity at the highest point of its trajectory has a magnitude of approximately 55.9 m/s and is directed at an angle of approximately 38.5° counterclockwise from the +x-axis.

Moving on to part (b) of the problem, we need to find the straight-line distance from where the projectile was launched to where it hits its target. To do this, we can use the horizontal component of the velocity and the time of flight.

The horizontal distance traveled by the projectile can be calculated using the equation:

D = Vix * t

Plugging in the values, we have:
D = (59.0 m/s) * (3.80 s)

Calculating this, we find:
D ≈ 224 m

Therefore, the straight-line distance from where the projectile was launched to where it hits its target is approximately 224 meters.

To solve this problem, we can break down the initial velocity of the projectile into its horizontal and vertical components.

The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the acceleration due to gravity.

Given:
Initial speed (V0) = 59.0 m/s
Launch angle (θ) = 32.0°
Time of flight (t) = 3.80 s

(a) Velocity at the highest point of the trajectory:
To find the velocity at the highest point, we need to determine the vertical component (Vy) at that point.

Vy = V0 * sin(θ)
Vy = 59.0 m/s * sin(32.0°)

Now, the velocity at the highest point will have the same magnitude as the initial velocity, but in the opposite direction. So the magnitude of the velocity at the highest point is 59.0 m/s.

The direction can be determined by finding the angle counterclockwise from the +x-axis. The angle at the highest point is 90° - θ.

Angle at highest point = 90° - 32.0° = 58.0° counterclockwise from the +x-axis.

Therefore, the projectile's velocity at the highest point of its trajectory has a magnitude of 59.0 m/s and a direction of 58.0° counterclockwise from the +x-axis.

(b) Straight-line distance from launch to the target:
To find this distance, we need to find the horizontal displacement of the projectile during the given time of flight (t).

The horizontal displacement (dx) can be calculated using the formula:

dx = Vx * t
Vx = V0 * cos(θ)

Vx = 59.0 m/s * cos(32.0°)
dx = (59.0 m/s * cos(32.0°)) * 3.80 s

Evaluate this expression to find the horizontal displacement (dx), which gives us the straight-line distance from where the projectile was launched to where it hits its target.