Hello! If anyone can help me answer these two questions with explained steps. Pl

Question

Hello! If anyone can help me answer these two questions with explained steps. Please and Thank you!

6. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

7. A 113 gram pear falls from a pear tree branch 3 m above the ground. How much time elaspsed before the pear hits the ground. With what speed the pear travels as it touches the ground?

Explanation / Answer

6. (a) The displacement of the 0bject is

s = u*t + (1/2)*a*t^2

= 0*1 + 0.5*9.8*1^2

= 4.9 m

(b) From the equation of motion

v^2 = u^2 + 2*a*s

The final velocity of the object is

v = sqrt(u^2 + 2*a*s)

= sqrt(0 + 2*9.8*75)

= 38 m/s

The final velocity will be negative as object is falling downwards.

(c) From the equation of motion is

y-y0 = u*t + 0.5*a*t^2

y = 0.5*a*t^2

The total time to fall is

t = sqrt(2*y/a)

= sqrt(2*75/9.8)

= 3.9 s

The displacement of the object in first 2.9 s i

s = u*t + (1/2)*a*t^2

= 0*2.9 + 0.5*9.8*2.9^2

= 41 m

The distance traveled by object in last 1 s is

D = 75-41

= 34 m

7. The time elaspsed before the pear hits the ground is

s = u*t + (1/2)*a*t^2

3 = 0*t + 0.5*9.8*t^2

t = 0.61 s

The speed the pear travels as it touches the ground is

v = u - g*t
  
= 0 - 9.8*0.61

v = -5.978 m/s

The speed of the pear before it hits ground is 5.978 m/s.

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