Plz answer this and brainliest i will give. dont skip! a ball is thrown horizontally from the top of a building 89 m high. the ball strikes the ground 80 m horizontally from the point of release. what is the speed of the ball just before it strikes the ground? answer in units of m/s.

Vox = ?

Voy = 0 m/s

g = 9.8 m/s

s = 61.7 m

h = 42.4 m

(1)...

To find the time taken for the ball to travel to the bottom:

For constant acceleration,

s = Voy*t + 0.5*g*t^2

42.4 = 0 + 0.5(9.8)*t^2

t^2 = 42.4 / 4.9

t^2 = 8.6531

t = 2.9417s

(2)...

For the intial velocity of the horizontal component (Vox) of the ball:

s = Vox*t + 0.5*a*t^2

There is no force acting on the horizontal component, so there is no acceleration.

s = Vox*t

61.7 = 2.9417*Vox

Vox = 181.5029 m/s

(3)...

Since there is no acceleration acting on the horizontal component, x, it remains constant throughout.

Hence, it is still 181.5029 m/s.

For the final velocity of the vertical component (Vfy) of the ball:

(Vfy)^2 = (Voy)^2 + 2*a*h

Acceleration in this case is the force of gravity.

(Vfy)^2 = 0 + 2*(9.8)*(42.4)

(Vfy)^2 = 831.04

Vfy = 28.8278 m/s

I've clearly explained every step. Hope that answers your question! =D

the speed of the ball just before it will be equal to 41.75 m/s

Explanation:

data provided:

h = height of building = 89 m

L = horizontal distance = 80 m

The speed of the ball at maximum height is assumed to be equal to zero. If we use the equation of motion we have:

s = u*t + (a*t^2)/2, where s = distance; u = initial velocity; a = gravity acceleration; t = time

replacing values and clearing t:

t = ((89*2)/9.8)^1/2 = 4.26 s

the speed will be equal to:

vy = uy + a*t = g*t = 9.8*4.26 = 41.75 m/s

what was the answer?

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