Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Answer:
maximum height = 5.1 m distance traveled/horizontal displacement = 35.3m
Explanation:
Vy = sin 30º • 20 m/s
Vy = 10 m/s
Vx = cos 30º • 20m/s
Vx = 17.3
V final = V initial + acceleration • time
0 = 10m/s + (-9.81 m/s/s) • time
subtract 10 m/s from both sides
-10m/s = -9.81m/s/s • time
divide each side by -9.81m/s/s
t = 10/9.81 s or roughly 1.02 second
the time to reach max height is equal to 1.02 s
Height = 1/2 • acceleration • time^2 + Vy • time
Height = 1/2 • (-9.81 m/s/s) • 1.02s^2 + 10m/s • 1.02 s
Height = 5.1 m
the maximum height of the ball is equal to 5.1m
since the ball starts from an initial height of zero and has a final height of 0 once it reaches the ground, the time needed to reach max height is the same amount of time to fall back down to earth. thus, time total is equal to 1.02s multiplied by 2
t total = 1.02s • 2
t total = 2.04s
horizontal displacement = Vx • time total
horizontal displacement = 17.3m/s • 2.04 s
Horizontal displacement = 35.3m
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
