Question

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.

Answer 1

A. xmax = 131.49 m

B. t = 8.74 s

C. ymax = 220.33 m

Step-by-step explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

`y=y_o+v_osin\theta-(1)/(2)gt^2` (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

`0=200+(25)sin53\°t-(1)/(2)(9.8)t^2\\\\0=200+19.96t-4.9t^2` (2)

You use the quadratic formula to obtain the value of t:

`t_(1,2)=(-19.96\pm√((19.96)^2-4(-4.9)(200)))/(2(-4.9))\\\\t_(1,2)=(-19.96\pm65.71)/(-9.8)\\\\t_1=8.74s\\\\t_2=-4.66s`

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

`x_(max)=v_ocos\theta t`

`x_(max)=(25m/s)(cos53\°)(8.74s)=131.49m`

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

`y_(max)=y_o+(v_o^2sin^2\theta)/(2g)` (3)

By replacing in the equation (3) the values of all parameters you obtain:

`y_(max)=200m+((25m/s)^2(sin53\°)^2)/(2(9.8m/s^2))\\\\y_(mac)=200m+20.33m=220.33m`

The maximum height reached by the cannon ball is 220.33m