a projectile is shot up 25 m/s at an angle of 34 degress how high imto the air does the projectile go?

To determine how high the projectile goes, we need to analyze its vertical motion.

Step 1: Split the initial velocity into horizontal and vertical components:
The initial velocity of 25 m/s at an angle of 34 degrees can be split into horizontal and vertical components using trigonometry. The vertical component (Vy) is calculated by multiplying the initial velocity (25 m/s) by the sine of the angle (34 degrees):

Vy = 25 m/s * sin(34 degrees)

Step 2: Calculate the time taken to reach the maximum height:
In a projectile's vertical motion, the vertical velocity decreases until it reaches zero at the maximum height. So, we can use the formula of vertical motion to find the time taken to reach the maximum height (t):

Vy = gt - (1/2)gt^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation, we get:

(1/2)gt^2 - gt + Vy = 0

This is a quadratic equation, where a = (1/2)g, b = -g, and c = Vy. Using the quadratic formula:

t = [-b ± √(b^2 - 4ac)] / (2a)

Substituting the values, we can solve for t.

Step 3: Calculate the maximum height:
Once we have the time (t), we can calculate the maximum height (h). The vertical position (h) at the maximum height is given by:

h = Vyi * t - (1/2)g * t^2

Where Vyi is the vertical component of initial velocity (Vy) and g is the acceleration due to gravity.

With these steps, we can now calculate the maximum height the projectile reaches.