Question

The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last second of the paddle) what speed was it just before the impact

Answer 1

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

$h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2$

h = 593.50 m

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

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

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

$v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s$

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

$h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2$

h₁₁ = 103 m

c)

Now, we use first equation of motion for complete motion:

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

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

$v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)$

vf = 107.91 m/s