plz solve this qs for me with a good explanation Show all work. Draw diagrams an
ID: 1591882 • Letter: P
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plz solve this qs for me with a good explanation
Show all work. Draw diagrams and explain answers as needed. Partial credit is available. Falling objects subject to aerodynamic drag accelerate initially but then reach what is called a terminal velocity. The motion from this point on is at constant velocity. Consider the motion of an object that has reached a terminal velocity of 28.6 m/s (down) at a height of 230m. If x=0 (the origin) is at ground level and "up" is the positive direction: Write out x_f =x_i + v_i t for this motion leaving x and t as variables. The data given represents the t = 0 motion state. Use the equation formulated in part (a) to determine the t value at which the object is 50 m high, and the t value at which the object hits the ground.Explanation / Answer
When the object is at the height of the 230 m from the ground it has achieved the terminal velocity.
hence the motion will be with constant velocity.
NOw Xf is the final position or distance from the ground at any time t
Xi is the initial distance from the ground at time t= 0 i.e when object achives terminal velocity
therefore Xi = 230 m
Now as object moving is downward therefore the distance from the ground will keep on decreasing.
Hence We can say in time t it will cover distance 28.6*t where 28.6 m/s is velocity
Now we can write final position at any time t
Xf = Xi - 28.6*t
Xf = 230 + (-28.6*t)
(b) When the object is at 50 m high that means Xf = 50 m
so putting the value in above equation
50 = 230 -28.6*t
t = 6.294 seconds
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