## Calculus Application Problem1

In this page calculus application problem1 we are going to see solution of the first problem of the topic application problems in calculus.calculus application problem1

Question 1:

A missile fired from ground level rises x meters vertically upwards in t seconds and x = 100t-25t²/2. Find (i) the initial velocity of the missile ,(ii) the time when the height of the missile is a maximum (iii) the maximum height reached (iv) the velocity with which the missile strikes the ground.

Solution:

Here the variable "x" represents distance taken by the missile and "t" represents times in seconds

The distance covered by the missile in "t" seconds

x = 100t-25t²/2

(i) The initial velocity of the missile

To find the initial velocity,we have to find the value of dx/dt, for that we have to apply 0 for t.

to find dx/dt let us differentiate the given function with respect to "t"

dx/dt = 100 (1) - 25 (2t)/2

= 100 - 25 t

put t = 0,to find the initial velocity

= 100 - 25 (0)

= 100 meter/seconds

(ii) the time when the height of the missile is a maximum

when the missile reaches its maximum height, then velocity of the missile will become zero.

that is dx/dt = 0

100 - 25 t = 0

- 25 t = -100

t = 100/25

t = 4 seconds

(iii) the maximum height reached

from the answer of the previous question we come to know that the missile reaches its maximum height in 4 seconds. From this to find the maximum height we have to apply t = 4 in the given equation. Because we need to find the value of x.

x  = 100 (4) - 25 (4)²/2

= 400 - 25 (16)/2

= 400 - 25 (8)

= 400 - 200

= 200 meter

(iv) the velocity with which the missile strikes the ground

when the missile strikes the ground,the distance will be zero.

100 t - 25 t²/2 = 0

(200 t - 25 t²)/2 = 0

200 t - 25 t² = 0

25 t (8 - t) = 0

t = 0 ,  t = 8

t = 0 is not possible.So let us take t = 8

velocity at t = 8

dx/dt = 100 - 25 t

= 100 - 25 (8)

= 100 - 200

=  -100 meter/second   calculus application problem1