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Masteriza [31]
1 year ago
6

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw

ay from and below the point of release. What is the magnitude of its velocity just before it strikes the ground?​
Physics
1 answer:
astraxan [27]1 year ago
5 0

26.54 m/s  is the magnitude of its velocity just before it strikes the ground

h=100m,v=20m/s,g=9.8m/s

time it takes to reach the ground,

[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]

x= 120m

t= 4.52

v= x/t

v= 120/4.52

v= 26.54 m/s

The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.

To know more about  velocity visit : brainly.com/question/16379705

#SPJ9

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The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter
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The value of normal force as the slider passes point B is

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The value of h when the normal force is zero

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<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

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Using summation of forces at B

Normal force, N  = ma + mg

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N = m(5gR/R + g)

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for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

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at C

for normal force to be zero

ma = mg

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v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

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3gR = 2gh

3R/2 = h

Learn more about normal force at:

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