Question

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range?

Answer 1

a)
`v_0=640` ft/s

b)
`y=H+1600` ft where
`H` represents the height as the projectile was launched.

c)
`x=11085.13` ft

Step-by-step explanation:

First, recognize the values that are given in the problem:

`\alpha =30 ^o`

`t_f=20`

`y_f=H`

a) With those three use this formula:
`y=H+v_0sin(\alpha)t-(1)/(2)gt^2`

to find the initial velocity
`v_0`.

`\\y_f= H+v_0sin(\alpha)t_f-(1)/(2)gt_f^2 \\H= H+v_0sin(30)(20)-(1)/(2)(32)(20)^2 \\ -v_0sin(30)(20)=-(1)/(2)(32)(20)^2\\ v_0=((1)/(2)(32)(20)^2)/(sin(30)(20)) \\ v_0=(6400)/(10) =640`

b) In order to find the maximum altitude, the time is needed to apply the formula. The maximum altitude is when the velocity in the y-axis is equal to zero, so use the formula for the velocity in the y-axis is to find the time, the formula is:

`v_y=v_(0y)-gt\\v_y=v_0sin(\alpha)-gt\\0=(640)sin(30)-32t\\32t=(640)sin(30)\\t=((640)sin(30))/(32)=(320)/(32)=10`

With
`s=10`s use this formula for the altitude:
`y=H+v_0sin(\alpha)t-(1)/(2)gt^2`

`y=H+v_0sin(\alpha)t-(1)/(2)gt^2\\y=H+(640)sin(30)(10)-(1)/(2)(32)(10)^2\\y=H+3200-1600\\y=H+1600`

Finally, the range is the maximum displacement in the x-axis, the formula of the displacement is:

`x=v_(0x)t=v_0cos(\alpha )t`

And the maximum occurs when
`t=20`s

`x=v_0cos(\alpha)t\\x=640cos(30)(20)\\x=11085.13`