Question

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

leaves the cannon 2m off the ground and lands in a net 1m off the ground. What height does she reach?
How much ground distance does she cover?

Answer 1

a

The height is
`H  = 6.74 \  m`

b

The horizontal distance is
`D  = 23.74 \  m`

Step-by-step explanation:

From the question we are told that

The speed is
`v  =  15 \  m/s`

The angle is
`\theta  =  40^o`

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

`H  =  (v^2 sin^2 \theta )/(2 *  g )  +  h`

=>
`H  =  ((15)^2 [sin (40)]^2 )/(2 * 9.8)  +  2`

=>
`H  = 6.74 \  m`

Generally from kinematic equation

`s = ut + (1)/(2) at^2`

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

`u  =  v sin (\theta)`

So

`-1 = (vsin 40 )t + (1)/(2) * (-9.8) t^2`

=>
`-4.9t^2 + 9.6418t + 1 = 0`

using quadratic formula to solve the equation we have

`t  =  2.07 \  s`

Generally distance covered along the horizontal is

`D  =  v cos (40) *  2.07`

=>
`D  =  15 cos (40) *  2.07`

=>
`D  = 23.74 \  m`