An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminus0.9tsquaredplus36t meters in t seconds. a. determine the velocity v of the object after t seconds. b. when does the object reach its highest point? c. what is the height of the object at the highest point? d. when does the object strike the ground? e. with what velocity does the object strike the ground? f. on what intervals is the speed increasing?

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

Differentiating with respect to time we get

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

b) The object will reach the highest point after 20 seconds

c) Highest point the object will reach is 360 m

d) Time taken to strike the ground would be 20+20 = 40 seconds

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

answersysvol bloat;