Question

Q2 (a) . A dynamite blast blows up a heavy rock with launch velocity of 160m/sec reaches a hight of s=160t-16t^2 ft after t sec, How high does the rock go
Find the velocity and speed of the rock when it is 256 ft above the ground on the way up and down
find the acceleration of the rock at time 5sec

Answer 1

Answer:

Maximum Height = 400m

Velocity and Speed = 144m/s

Acceleration= -32m/s²

Step-by-step explanation:

As we are given:

The launch velocity or simply the initial velocity = vi = 160m/s

And,

Displacement as a function of time = s = 160t - 16t²

And we are required to find:

The maximum height it will achieve with that velocity= Smax

The velocity and speed at a height of 256m above the ground level= S

The acceleration at t = 5s = a.

For maximum height:

To find the maximum height achieved, we must know how much time will it take to reach it's maximum height;

And for that we use the extremas of a function by derivatives:

Now, differentiating the given function with respect to time;

ds/dt = d/dt(160t-16t²)

ds/dt = 160-32t

And, for extreme value(maximum in our case for displacement), the derivative(ds/dt) should be '0';

0 = 160-32t

t = 160/32

t = 5s.

So, putting the value of time at which the displacement(height) would be maximum:

s = 160(5)-16(5)²

s = 400m

Or, for naming convention;

Smax = 400m.

Velocity and Speed when it is 256ft high:

Firstly, converting 256ft into meters;

1ft = 0.3 meters

Thus,

256ft = 0.3 × 256 meters

256ft = 76.8 meters

Now, for the velocity, we must know how much time it would take to reach at this height, and again for that, we use the given function:

s = 160t-16t²

76.8 = 160t - 16t²

16t²-160t+76.8= 0

And that's a quadratic equation, it's roots by solving would be:

t1 = 9.49s and t2 = 0.5 s

Now which one to pick?

Obviously, 't2' as t1 is greater than the total time taken to reach the maximum height

Thus, to calculate velocity at 't2= 0.5s':

We know that velocity is nothing but the derivative of displacement;

ds/dt = v = d(160t-16t²)/dt

v = 160-32t --(A)

And for that velocity, put t=t2= 0.5s

v = 160-32(0.5)

v = 144m/s.

And we know that speed is the magnitude of velocity, so speed is also '144m/s'.

Acceleration at t=5s:

As we know that acceleration is the rate of change(derivative) of velocity and velocity is itself the derivative of displacement; therefore we can use eq(A);

v = 160-32t

And,

a = dv/dt

a = -32.

Thus, the acceleration remains constant for any time.