Question

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the ground with a speed of 11 m/s, how long was he in the air (in seconds)? 8. If a body travels half its total path in the last 1.10 s of its fall from rest, find the total time of its fall (in seconds). 9. An arrow is shot straight up in the air with an initial speed of 240 ft/s. If on striking the ground it embeds itself 8.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow, in units of feet/second 2.

Answer 1

3.49 seconds

3.75 seconds

-43200 ft/s²

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

`s=ut+(1)/(2)at^2\\\Rightarrow 50=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(50* 2)/(9.81)}\\\Rightarrow t=3.19\ s`

Time the parachutist falls without friction is 3.19 seconds

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 9.81* 50+0^2)\\\Rightarrow v=31.32\ m/s`

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

`v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=(11-31.32)/(-67)=0.3\ s`

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow (s)/(2)=0t+(1)/(2)* a* t^2\\\Rightarrow (s)/(2)=(1)/(2)at^2`

`s=ut+(1)/(2)at^2\\\Rightarrow (s)/(2)=u1.1+(1)/(2)* a* 1.1^2`

Now the initial velocity of the last half height will be the final velocity of the first half height.

`v=u+at\\\Rightarrow v=at`

Since the height are equal

`(1)/(2)at^2=u1.1+(1)/(2)* a* 1.1^2\\\Rightarrow (1)/(2)at^2=at1.1+(1)/(2)* a* 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0`

`t=(11\left(1+√(2)\right))/(10),\:t=(11\left(1-√(2)\right))/(10)\\\Rightarrow t=2.65, -0.45`

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-240^2)/(2* (8)/(12))\\\Rightarrow a=-43200\ ft/s^2`

Magnitude of acceleration is -43200 ft/s²