Question

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?

Answer 1

Answer:

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:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

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

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

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

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

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

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

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 $s=ut+\frac{1}{2}at^2$

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

A) The velocity v of the object after t seconds is; v(t) = -1.8t + 36

B) The time it will take the object to reach its highest​ point is; t = 20 s

C) The height of the object at the highest​ point is; 360 m

D) The time it takes for the object strike the​ ground is; 40 s

E) The velocity at which the object strikes the​ ground is; 36 m/s

F) The interval on the speed increases is; from time of t = 20 seconds to t = 40 seconds.

We are given the equation of height as;

s = -0.9t² + 36t

• A) Since we have height equation, we know that velocity is a derivative of height distance. Thus;

Velocity; v(t) = ds/dt = -1.8t + 36

v(t) = -1.8t + 36

• B) The object will get to the highest point when velocity is zero.

Thus; at v(t) = 0 m/s, we have;

0 = -1.8t + 36

Rearranging gives us;

1.8t = 36

t = 36/1.8

t = 20 s

• C) Since the object gets to the highest point after 20 s, it means the height at this point will be;

s(20) = -0.9(20)² + 36(20)

s(20) = -360 + 720

s(20) = 360 m

D) Since it took a time of 20 seconds to get to its' highest point, it means it will take the same time to get to the ground from that highest point.

Thus; Total time to strike the ground = 20 + 20 = 40 s

E) To find the velocity with which the object will strike the ground, we will make use of newton's first equation of motion;

v = u - at

Since it took 20 seconds from highest point till it hits the ground, then t = 20 s

Now, from earlier; v(t) = -1.8t + 36. However, acceleration is the derivative of velocity. Thus; a = dv/dt = -1.8

Thus;

v = 0 - (-1.8 × 20)

v = 36 m/s

• F) From our calculations, we see that the speed is decreasing from when you throw the object because at the highest point it is 0 m/s, however when it is going to the ground, it increases to 36 m/s.

Thus, the interval that the speed increases is from t = 20 seconds to t = 40 seconds.

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