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A hiker throws a ball at an angle of 21.0° above the horizontal from a hill 21.0 m high. The hiker’s height is 1.750 m. The magnitudes of the horizontal and vertical components of the velocity are 14.004 m/s and 5.376 m/s, respectively. Find the distance between the base of the hill and the point where the ball hits the ground. (Consider the hiker’s height while calculating the answer.)

Respuesta :

Answer:

38.8 m

Explanation:

h = -(21 + 1.75) = - 22.75 m

g = - 9.8 m /s^2

Ux = 14.004 m/s

Uy = + 5.376 m/s

Let the ball hits the ground in time t and at a distance d from the base of hill.

Use second equation of motion

h = Uyt + 0.5 at^2

- 22.75 = 5.376 t - 0.5 x 9.8 t^2

4.9 t^2 - 5.376 t - 22.75 = 0

[tex]t=\frac{5.376\pm \sqrt{5.376^{2}+4\times 4.9\times 22.75}}{9.8}[/tex]

By solving

t = 2.77 second

So, horizontal distance

d = Ux t

d = 14.004 x 2.77 = 38.8 m

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