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A 0.65 kg rock is projected from the edge of
the top of a building with an initial velocity of
10.5 m/s at an angle 40◦
above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 19.6 m from the base
of the building. How tall is the building? Assume the
ground is level and that the side of the building is vertical. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m

A 065 kg rock is projected from the edge of the top of a building with an initial velocity of 105 ms at an angle 40 above the horizontal Due to gravity the rock class=

Respuesta :

samuelonum1

Answer:

180.45m

Explanation:

Given that

mass m= 0.65kg

let H be height of building

angle of projection =40°

initial velocity  v up = 10.5 sin 40 = 6.75 m/s

horizontal velocity u= 10.5 cos 40 = 8.04 m/s

horizontal distance x = 19.6

x=ut

x= u t =

x= 8.04 t

t=19.6/8.04  = 2.44 seconds

applying the formula

h = H + Vt -1/2gt^2

0 = H + 6.75 * 2.44 - 4.9 * 2.44^2

solving for H we have

0=H + 6.75 * 2.44 - 4.9 * 5.9536

0=H + 6.75 * 2.44 - 29.17264

0=H + 6.75 *-26.73264

0=H -180.44532

H=180.45m

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