Question

On level ground a shell is fired with an initial velocity of 34.0 m/s at 51.0° above the horizontal and feels no appreciable air resistance. A. Find the horizontal and vertical components of the shell's initial velocity.

B. How long does it take the shell to reach its highest point?
C. Find its maximum height above the ground.
D. How far from its firing point does the shell land?
E. At its highest point, find the horizontal and vertical components of its acceleration.
F. At its highest point, find the horizontal and vertical components of its velocity.

Answer 1

A.
`v_(x)=21.4m/s`

`v_(y)=26.42m/s`

B.
`t=2.7s`

C.
`y_(max) =35.61m`

D.
`x=115.34m`

E.
`a_(x)=0`

`a_(y)=-9.8m/s^2=g`

F.
`v_(y)=0`

`v_(x)=21.4m/s`

Step-by-step explanation:

From the exercise our initial values are:

`v_(o)=34m/s`

`\alpha=51º`

A. The horizontal and vertical components are:

`v_(x)=34cos(51)=21.4m/s`

`v_(y)=34sin(51)=26.42m/s`

B. At maximum height the y-component of velocity becomes 0

`v_(y)=v_(o)+a_(y)t`

`0=26.42-9.8m/s^2*t`

`t=2.7s`

C. The maximum height above the ground is:

`v_(y) ^(2)=v_(o)^2+2a_(y)(y-y_(o))`

At maximum height the y-component of velocity becomes 0

`0=(26.42)^2-2(9.8)y`

`y=(-(26.42)^2)/(-2(9.8))=35.61m`

D. To find how dar from its firing point does the sell land we need to calculate how much time does it take to do it first

`y=y_(o)+v_(oy)t+(1)/(2)gt^2`

When the shell land y=0

`0=0+(26.42)t-(1)/(2)(9.8)t^2`

Solving the quadratic equation for t

`t=0` or
`t=5.39s`

Since time can not be 0 t=5.39s

`x=v_(ox)t=(21.4m/s)(5.39s)=115.34m`

E. Since the velocity at the horizontal component is constant

`a_(x)=0`

The vertical acceleration of the shell is gravity

`g=-9.8m/s^2`

F. At highest point the vertical component is 0. The shell stops going up ans start to go down

`v_(y)=0`

`v_(x)=v_(oy)+a_(x)t`

Since
`a_(x)=0`

`v_(x)=21.4m/s`