You are standing near a building and throw a ball straight up with an initial sp

Question

You are standing near a building and throw a ball straight up with an initial speed of v. The ball rises to a highest point at half the building's height. You now wish to throw the ball so that it just rises to the top of the building. What should be the initial speed of the ball in the second case? A ball is shot vertically upward from the surface of a planet in a distant solar system. The acceleration due to gravity on this planet is not known, but can be figured out from the observation that the ball reaches a maximum height of 25 m and 'remains in the air' (for its complete up and down motion) for 5 seconds. What is the magnitude of the acceleration due to gravity on this planet? What is the magnitude of the initial velocity of the ball?

Explanation / Answer

2.3.2

let us say the height of the building is h

then in case 1) h/2 = v^2/(2*g)

h = v^2/g

now for case 2)

h = vx^/(2*g)

then v^2/g = vx^2/(2*g)

vx = sqrt(2)*v is the required initial speed of the ball to reach top of the building

2.3.3

maximum height Hmax = U^2/(2*g')

g' is the accelaration due to gravity on this new planet

Hmax = 25 m

and time of flight T = 2*U/g' = 5

Hmax /T = (U^2/(2*g'))*(g'/(2*U)) =

U/4 = 25/5

initial velocity is U = 100/5 = 20 m/s

accelaration due to gravity is g' = (2*U)/5 = (2*20)/5 = 8 m/s^2

So the answers are a) 8 m/s^2

b) U = 20 m/s

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.