<span>A ball is thrown at a 30 degree angle above the horizontal with a speed of 10 ft/s. After 0.50s the horizontal component of it's velocity will be the same. In a projectile motion the horizontal component of the velocity is said to be constant. Therefore, it will be equal to the initial velocity.</span>
162
+207
= 369
Okay, so it says he descends 285 feet after he's climbed 369 feet. That means he needs to descend a bit more to reach 453 feet, which we'll just call 0 for now. First, we need to do some subtraction:
369
- 285
= 84
Okay, we know he needs to descend 84 feet to get to his starting point.
So your answer is C
Wait, could you please restate what I'm supposed to answer xD Thank you
<em>PQR with vertices P(–2, 9), Q(7, –3), and R(–2, –3)</em>
<em>first distance P(–2, 9), Q(7, –3) </em>
<em>The distance (d) between two points is given by the following formula: </em>
<em>Answer= 15</em>
