Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?

Sagot :

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

OlanmaE

Answer:

0.774m

Explanation:

The formula for maximum height is given by:

hmax = ∨₀² ₓ Sin (α)² / 2 × g

where;

∨₀ = initial velocity

Sin (α) = angle of launch

g = gravitational acceleration which is equal to 9.8m/s²

Plugging in our values, we will have:

hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²

hmax= 20.25m/s × 0.75 / 19.8m/s²

hmax = 15.1875 / 19.8

hmax = 0.774m

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.