- #1

- 22

- 0

## Homework Statement

A projectile has a time of flight 7.5s and a range 1200m

calculate

(a) its horizontal velocity

(b) Its maximum height

(c) the velocity witch which it is produced

I take a

_{y}as -9.8ms

^{-2}

## Homework Equations

r = u

_{x}t

d

_{y}= u

_{y}t + (a

_{y}t

^{2}/2)

v

_{y}

^{2}= u

_{y}

^{2}+ 2ah

u

_{x}

^{2}+ u

_{y}

^{2}= u

v = u + at

## The Attempt at a Solution

I hope I am the correct

a) r = u

_{x}t

u

_{x}= r/t

u

_{x}= 1200/7.5

u

_{x}= 160m

.: Initial Horizontal Velocity is 160m

b)

d

_{y}= u

_{y}t + (at

^{2}/2)

d

_{y}= 0 (Total vertical displacement is zero)

0 = 7.5u

_{y}- 275.625

7.5u

_{y}= 275.625

u

_{y}= 36.75

v

_{y}

^{2}= u

_{y}

^{2}+ 2ah

v

_{y}= 0

0 = 1350.5625 - 19.6h

19.6h = 1350.5625

h = 68.90625 ...

.: Maximum height is 68.91m

^{-1}

c)

u

_{x}

^{2}+ u

_{y}

^{2}= u

(160)

^{2}+ (36.75)

^{2}= 26950.5625

u = 164.1662648 ...

.: Initial Velocity is 164.17ms

^{-1}

v = u + at

v = (164.1662648) + (-9.8)(7.5)

v = 90.6662648 ...

.: Final Velocity is 90.67ms

^{-1}

Ya ?