Aparachutist 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.

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Time the parachutist falls without friction is 3.19 seconds

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

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

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

Since the height are equal

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.

Magnitude of acceleration is -43200 ft/s²

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west