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

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft