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Free Fall Take g = 9.81, Neglect air resistance in the following calculations Assume an object is thrown upwards off a 50.0my

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7. The formula for time taken to reach the top is the time of ascent

Given the body thrown upwards, so initial velocity is given as Vi = 12 m/s and final velocity for such a body is Uy 0 m/s.

Time of ascent is given by ひ 12 ta =ー== 1.223 s 9.81 .

The closest option is Option (b) which is the answer.

8. The initial distance = 50.0 m.

So the total distance that the body travels can be given from the formula y=y_i+v_it+rac{1}{2}at^2 .

Here t=t_a. So we get

y = 50 + 12 × 1.22 + 0.5 × (-9.81) (1.22)50 14.64-7.30 57.34 m2.

Which gives the closest answer Option(a). So the answer is Option(a).

9. The answer is the total time of flight. Here the body initially is at a height 50 m. So first let us calculate the time for the body to go from height 50 m and come back. So the time taken by the body is the time taken to reach maximum height and come back down to the ground. Let us visualize the motion as two parts. One time taken to reach maximum height. i.e. 1.22 s the time of ascent.

Two the time taken by the body to fall back from a height of 57.34 m to the ground.

So we have from equation of motion

Delta y=v_it+rac{1}{2}at^2

Here ▲y-57.34 m , a = g =9.81 m/s v_i=0 as the body falls down from maximum where the velocity is zero.

so we have 57.34 = 0.5 × 9.81 × t2

from this 57.34 0.5 x 9.81 11.69 s2,t-V11.69- 3.419 s

So the total time = t_{total}=3.419 s + 1.22 s = 4.63 s

The closest option is Option (c). So the correct answer is Option (c).

10. The speed can be calculated from the formula v_f-v_i=at

Here v_i=0, a=g=9.81 m/s^2, t=3.419 s

So we have vf -0- 9.81 x 3.419-33.54 m/s .

The answer is Option (c).

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