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3 years ago

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A 0.75 kg rock is projected from the edge of the top of a building with an initial velocity of 11.9 m/s at an angle 59◦ above th

e horizontal. The building is 14.2 m in height. At what horizontal distance, x, from the base of the building will the rock strike the ground? 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.

1 answer:

Vesna [10]3 years ago

4 0

Answer:

12.76 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 11.9 m/s

Angle of projection (θ) = 59°

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (i.e Range) (R) =?

The horizontal distance other wise known as range from the base of the building to which the rock will strike the ground can be obtained as follow:

R = u²Sine2θ / g

R = 11.9² × Sine (2×59) / 9.8

R = 141.61 × Sine 118 / 9.8

R = 12.76 m

Therefore, the horizontal distance from the base of the building to which the rock will strike the ground is 12.76 m

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$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

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We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

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We are given:

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Putting values in equation 2, we get:

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We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

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