3. A projectile is fired from ground level at an angle of 30.0° and a speed of 20.0 m/s. The projectile is fired at a wall 25.0 m away. How far up on the wall will the projectile hit?

Start by finding the time interval, delta t, that it takes for the item to hit the wall. You can do so by knowing that t = d/s. We can solve for s, which is speed, by finding the initial velocity in the horizontal direction.

We can find this by using the launch angle given, 30 degrees, and solving for its horizontal component (parallel to x-axis).

This is found by cos30 = v_x/20

Solve for v_x to get: 17.32051 m/s. Now we can use this initial velocity in the horizontal direction in the equations.

Time = distance/speed

Time = 25 m/17.32051 m/s = 1.44338 seconds

Now we have t = 1.44338 seconds

We want to find the height of the item at a certain time, this time is 1.44338 seconds. We don't have the final velocity, so therefore, we can use the kinematic equation for constant acceleration (object in projectile motion has constant acceleration of -9.8 m/s^2 due to gravity):

No Final Velocity

x_f = x_i + v_i * t + 1/2a (t^2)

x_f and x_i are final and initial positions, respectively, and v_i is the initial velocity. The initial position of this item is 0 m, since we are starting from ground level.

We are trying to solve for the final position, x_f, and we know the variables v_i (17.32051 m/s), t (1.44338 s), and a (9.8 m/s).

Substitute these values into the equation to find the final position:

x_f = (0) + (17.32051 m/s)(1.44338 s) + 1/2(9.8 m/s^2)(1.44338 s)^2

x_f = (25.00007772) + (10.20839454)

x_f = 35.20847227

Now we know that at the time it takes for the item to hit the wall, 1.44338 s, the item's height is 35.20847 meters.

Let AB,BC and CA are the sides of triangle ABC, Given that, Let AB=15cm,BC=27cm and Third side CA=? Then, We know that

CA=BC−AB. CA=15−27.

all in all: CA=12cm