im stuck i have no clue where to start, please help me i just know the givens and some claculations.

question: kevin stands on he top of a buiulding and he throws an aboject up into the air at a velocity of 16.0m/s at an angle of 25 degrees to the horizontal. if the building is 75 m tall, how far from the base will the object hit the ground.

given:
vi: 16.o m/s
dy: 75m
dx:?
a=9.8m/s^2
calculated given:vix=6.76m/s,viy=14.5m/s

this is all i know, i have know clue what to do, please help me. Thanks!!

The vertical component viy should be LESS than the horizontal component, vix. That is because the launch angle is closer to vertical than horizontal.

Use the equation of motion that contains g and viy to compute the time that it takes for the object to hit the ground (y=0).

y = 75 + viy*t - (g/2)*t^2 = 0

Once you have solved for t (using the positive of the two possible solutions), use that and vix to compute the horizontal location.

X = vix*t

To solve this problem, you can use the kinematic equations of motion. Let's break down the question and the steps you can take:

1. Start by analyzing the motion of the object. The object is launched with an initial velocity of 16.0 m/s at an angle of 25 degrees to the horizontal. This means that there is an initial velocity in the horizontal direction (vix) and the vertical direction (viy).

2. Calculate the horizontal and vertical components of the initial velocity:
- The horizontal component (vix) is given by: vix = vi * cos(theta), where vi is the magnitude of the initial velocity (16.0 m/s) and theta is the launch angle (25 degrees). Substitute the given values to find vix.
- The vertical component (viy) is given by: viy = vi * sin(theta). Substitute the given values to find viy.

3. Now, consider the vertical motion of the object. The object is thrown up, reaches a maximum height, and then falls back down. The vertical motion can be analyzed using these equations:
- The equation for vertical displacement (dy) is: dy = viy * t + (1/2) * a * t^2, where dy is the vertical displacement (75 m), viy is the initial vertical velocity (calculated in step 2), a is the acceleration due to gravity (9.8 m/s^2), and t is the time taken for the object to reach the ground.

4. Since the object is thrown up and then comes back down, the final vertical displacement (dy) will be zero. Use this information to solve for the time taken (t).

5. With the time (t) calculated, you can now determine the horizontal position of the object at that time by using the equation: dx = vix * t, where dx is the horizontal displacement. Substitute the calculated values for vix and t.

These steps should help you solve the problem and find the horizontal displacement (dx) of the object from the base of the building when it hits the ground. Make sure to substitute the known values correctly into the equations and double-check your calculations.