(c) Upon choosi acceleration of the rock as functions of the time, t 10,2.50 s).

Question

(c) Upon choosi acceleration of the rock as functions of the time, t 10,2.50 s). ng an appropriate coordinate system, accurately sketch grapi moving vertically upward with a speed of 10.0 m/s. The pilot of the hot-air balloon gingerly oon at the instant that the balloon is 50.0 m above the ground, after which the sandbag is freely falling. (a) Measured relative to the ground, to what maximum height does the sandbag (b) Compute the position of the sandbag at 0.500 s and 2.00 s after being released. (c) For how much time is the sandbag in flight after being released? (d) How fast is the sandbag moving immediately before hitting the ground? (e) Upon choosing an appropriate acceleration of the sandbag as functions of the point immediately before it strikes the ground. coordinate system, accurately sketch graphs of the position, velocity, and the time, t, starting from when the sandbag is released up to from the top of a tall building. After it has been falling for a few seconds, it falls 80.0 m in interval. Upon ignoring air resistance, what distance will it fall during the next 2.00-s interval? 13. A brick is dropped a 2.00-s time U

Explanation / Answer

Let the motion of sand bag begin at t=0 secs

It starts upward flight with initial velocity = 10m/s and decelerartes with -9.8 m/s^2 for t seconds before it's velocity becomes zero at highest point in it's flight.

so 0 = 10m/s - 9.8 X t So t = 10/9.8 = 1.02 secs , for this time upward flight continues.

In this time of 1.02 secs ,the sand bag travels a distance = 10m/s X 1.02 sec - 0.5 X 9.8m/s^2 X (1.02secs)^2

= 5.10 m

So the highest point from ground level = 50m+5.10 m = 55.10m

(b) In this part we want to know where sand bag is at t= 0.500 secs

distance travelled upwards from initial position in 0.500 secs = 10m/sX 0.500secs-0.5 x 9.8 X 0.5^2

= 3.775m

Distance from bottom at t = 0.5 secs = 50+3.775 = 53.8m

We also want position at t= 2secs , with in these 2 seconds,1.02 secs is consumed in upward travel to a height of 55.10 and from there the sand bag begins downward journey for (2- 1.02)secs

So distance travelled downwards = 0.5 X 9.8 X (2-1.02)s^2 = 4.71 meters

So positon from ground = 55.1- 4.71 = 50.4 m

(c) Total time of flight = time for upward flight (Tup) + time for downward flight (Tdn)

time for upward flight = 1.02secs

55.10 mrs = 0.5 X 9.8 m/s^2 X (Tdn)^2 So Tdn = sqrt( 55.1 / 0.5X 9.8m/s^2) = 3.35 secs

Total time of flight = 1.02 secs + 3.35 secs = 4.37 secs

(d) In 1.02 secs highest point of 55.1m is attained where velocity of sand bag = 0

Thereafter it starts accelerating downwards up to ground level and if v is velocity at ground level,

v^2 = 2 X 9.8 m/s^2 X 55.1m

v= 32.9m/s

Note: I am allowed to solve single question with not more than 4 sub-parts.Sorry about that.

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