Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of v meters per second. Then v
dropped from a height of 42,6 meters, with what velocity does it hit the ground?
Round your answer to the pearest tenth.

28.9m/s

Step-by-step explanation:

The following were obtained from the question:

H = 42.6m

g = 9.8m/s2

To get the velocity, we must first obtain the time take for the object to fall from the height. This can be achieved by:

H = 1/2 gt^2

42.6 = 1/2 x 9.8 x t^2

42.6 = 9.8t^2 / 2

Cross multiply

9.8t^2 = 42.6 x 2

Divide both side by 9.8

t^2 = (42.6 x 2) / 9.8

Take the square root of both side

t = √[(42.6 x 2) / 9.8]

t = 2.95secs

Now we can obtain the velocity as follows:

V = gt

V = 9.8 x 2.95

V = 28.9m/s

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

group similar terms to get,

this implies that,

take positive square root of both sides to get,

this implies that,