Projectile motions :( (1 Viewer)

[MrLuvable]

1.A projectile has a time of flight of 7.5 s and a range of 1200m. Calculate:

a) its horizontal velocity

Δx = u(x) x t
1200=7.5u
u(x) = 160m/s

b) its maximum height
c) the velocity with which it is projected

2. A cannon ball is fired at 80ms-1 at an angle of 45 degrees to the horizontal. Calculate the height at which the ball hits a vertical cliff 150m away.

Thank youu !!

 

[anon09]

b) When a projectile reaches its maximum height, v = 0.

To find max. h, you can use either vy^2 = uy^2 + 2aysy or sy = uyt + 1/2ayt^2

However, both require knowledge of uy.

So, using sy = uyt + 1/2ayt^2:

0 = 7.5uy - 4.9 x 7.5^2

7.5uy = 275.625

uy = 36.75ms-1


Now to find max. h:

vy^2 = uy^2 +2aysy

0^2 = 36.75^2 - 19.6sy

19.6sy = 1350.5625

sy = 68.9m (maximum height)


c) You now have both ux and uy so you can find initial velocity by solving a pair of simultaneous equations.

ux = 160 = ucos@ and uy = 36.75 = usin@

Therefore, u = 160/cos@ and u = 36.75/sin@

Equate the two values: 160/cos@ = 36.75/sin@

160 sin@/cos@ = 36.75

160 tan@ = 36.75

tan@ = 36.75/160

@ = tan-1 (36.75/160) = 12degrees 56minutes 8.73seconds

u = 160/cos 12d 56m 8.73s

Hence, the intial velocity is approx. 164.17ms-1 at 13degrees to the horizontal.

 

[darkchild69]

1.A projectile has a time of flight of 7.5 s and a range of 1200m. Calculate:

a) its horizontal velocity

Δx = u(x) x t
1200=7.5u
u(x) = 160m/s

b) its maximum height
c) the velocity with which it is projected

2. A cannon ball is fired at 80ms-1 at an angle of 45 degrees to the horizontal. Calculate the height at which the ball hits a vertical cliff 150m away.

Thank youu !!

Seeing as the first q was done, i'll do the second one

Firstly, u need to split the vector into it's components, seeing as it is at 45 degrees, you should expect both ux and uy to be the same!

Ux = UCos45

Ux = 56.57ms^-1

Uy = USin45

Uy = 56.57ms^-1

Now, we need to figure out how long it will take to travel a horizontal displacement of 150m.

Sx = Ux*t
150 = 56.57*t
t = 150/56.57
t = 2.65s

Now, using this time, we need to figure out what the vertical displacement of the cannon ball would be after travelling the 150m!

So:

Sy = Uy*t +1/2ay(t)^2

Sy = 56.57*2.65 -4.9(2.65)^2

Sy = 150m

Therefore the cannon ball will hit the cliff 150m above where it was fired from

I hope that is the answer, i left all my calculators at work, had to rely on my phone