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

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Step-by-step explanation:

a)

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

`h = v_(i)t + ((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) + ((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 + ((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) + ((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