A particle is projected with a speed of 39.2m/sec at an elevation of 30degrees.find its greatest height.

at max height, v=0

sin 30 = 1/2, so the velocity is given by

39.2/2 - 9.8t = 0
t = 4

So, it rose for 4 seconds to max height.

Now recall that the height is given by

h = 39.2t - 4.9t^2

Just plug in t=4

How did you get 9.8t pls

To find the greatest height reached by the particle, we need to analyze the motion of the particle in two separate components: horizontal and vertical.

First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, unaffected by gravity. The vertical component (Vy) changes due to the acceleration of gravity (-9.8 m/s^2).

Given:
Initial speed (V): 39.2 m/s
Launch angle (θ): 30 degrees

Horizontal Component:
To find Vx, we use the formula: Vx = V * cos(θ).
Vx = 39.2 m/s * cos(30 degrees)
Vx ≈ 33.94 m/s

Vertical Component:
To find Vy at the highest point, we first need to find the time taken to reach that point. We use the formula: Vy = V * sin(θ) - gt, where g is the acceleration due to gravity (-9.8 m/s^2) and t is the time.
0 = 39.2 m/s * sin(30 degrees) - 9.8 m/s^2 * t

Rearranging the equation to solve for t:
t = (39.2 m/s * sin(30 degrees)) / 9.8 m/s^2
t ≈ 2 seconds

Now that we know the time (t), we can find the maximum height (H) using the formula: H = Vy * t + (1/2) * g * t^2
H = 39.2 m/s * sin(30 degrees) * 2 s + (1/2) * (-9.8 m/s^2) * (2 s)^2
H ≈ 19.6 m * 2 s - 9.8 m/s^2 * 4 s^2
H ≈ 39.2 m - 39.2 m
H ≈ 0 m

Therefore, the greatest height reached by the particle is approximately 0 meters.