A projectile is fired on level ground, and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. After 7 seconds of flight, what is the approximate angle of the projectile's velocity vector with respect to horizontal?

Question

A projectile is fired on level ground, and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. After 7 seconds of flight, what is the approximate angle of the projectile's velocity vector with respect to horizontal?

So I figured out the angle to the horizontal the projectile is fired at is 53 degrees, but I don't know much else to do :(

Answer 1

The vertical component is vy = 40-9.8t

so the angle θ at t=7 is tanθ = (40-9.8*7)/30 = -28.6/40
θ = 144.44°

Answer 2

Vo = 30+40i = 50[53o]

yo+g*T = 0
40-9.8T = 0
T = 4.1 s. to reach max ht.
Y = Yo+gT = 0+9.8*(7 -4.1) = 28.4 m/s.
Tan A = Y/X = 28.4/30
A = 43.5o

Answer 3

To find the angle of the projectile's velocity vector with respect to the horizontal, you can use inverse trigonometric functions. Here's how you can solve the problem:

1. Start by breaking down the initial velocity into its horizontal and vertical components. Given that the initial horizontal component of velocity (Vx) is equal to 30 m/s and the initial vertical component of velocity (Vy) is equal to 40 m/s.

2. The time of flight of the projectile is given as 7 seconds. Since the motion is symmetric, the time taken to reach the highest point will be half of the total time of flight, which is 7/2 = 3.5 seconds.

3. During this time, the vertical component of velocity (Vy) decreases due to the acceleration due to gravity (g), which is approximately 9.8 m/s^2. Therefore, the vertical component of velocity (Vy) at the highest point can be calculated using the equation:

Vy = Vy_initial - g * time
Vy = 40 m/s - 9.8 m/s^2 * 3.5 s

4. Next, calculate the magnitude of the projectile's velocity (v) at the highest point using the Pythagorean theorem. The magnitude of the velocity vector (v) at any point is given by:

v = sqrt(Vx^2 + Vy^2)

5. With the values of Vx and Vy at the highest point, substitute them into the equation above to get the magnitude of the velocity (v) at that point.

6. Finally, to find the angle of the projectile's velocity vector with respect to the horizontal, use the inverse tangent function (tan^(-1)):

angle = tan^(-1)(Vy/Vx)

Substitute the values of Vy and Vx at the highest point into the equation above to find the angle.

By following these steps, you can determine the approximate angle of the projectile's velocity vector with respect to the horizontal.