Question

. 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.

Answer 1

a) 4.9 m

b) 38.3 m/s

c) 33.4 m

What are three equation of motion?

The three equations are,

v = u + at

v² = u² + 2as

s = ut + ½at²

According to the question,

Let hold the y-axis in the direction of gravity.

Initial velocity 0 = 0 /,

Initial height ℎ0 = 0 , and finish height ℎ = 75.0

gravitational acceleration = 9.8 /²

a) Determine the distance traveled during the first second.

By third equation of motion

$S_{0} = v_{0} t +\frac{1}{2} gt^{2}$  where $v_{0}$ =0

then $S_{0} = \frac{gt^{2}}{2} = \frac{9.8*1^{2} }{2} = 4.9 m$

b) Determine the final velocity at which the object hits the ground

let – velocity of object, when it falling to the ground

– moment falling to the ground, and obtain from condition:

ℎ = ℎ() = 75.0

$v_{f} = v_{0} +gt = gt$

$h (t_{f} ) = \frac{gt^{2} }{2} = 75.0 m\\t_{f} =\sqrt{\frac{2h_{f} }{g} } \\v_{f}= gt = g \sqrt{\frac{2h_{f} }{g} } =\sqrt{2hg_{f} }= =\sqrt{2*9.8*75.0} = 38.3 m/s$

c) Determine the distance traveled during the last second of motion before hitting the ground.

From the above equation we get,

$\sqrt{2*75*9.8*1} - \frac{9.8*1^{2} }{2} = 38.3 -4.9 = 33.4 m$

Therefore,

a) 4.9 m is the distance traveled during the first second.

b) 38.3 m/s is the final velocity at which the object hits the ground

c) 33.4 is the distance traveled during the last second of motion before hitting the ground.

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