Projectile Motion Q from dot point physics (1 Viewer)

Neha

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heyy i need help with this question

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 calculated the horizontal velocity to be 160m/s but I dont know how to get the maximum height or the velocity.
 

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Neha said:
heyy i need help with this question

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 calculated the horizontal velocity to be 160m/s but I dont know how to get the maximum height or the velocity.
Let me have a crack at it ..

b)

dy = uyt + (at2/2)

dy = 0 (Total vertical displacement is zero)
0 = 7.5uy - 275.625
7.5uy = 275.625
uy = 36.75

vy2 = uy2 + 2ah

vy = 0
0 = 1350.5625 - 19.6h
19.6h = 1350.5625
h = 68.90625 ...

.: Maximum height is 68.91m




c)

ux2 + uy2 = u

(160)2 + (36.75)2 = 26950.5625
u = 164.1662648 ...

(c) the velocity witch which it is produced
witch velocity?
Well initial velocity is 164.17ms-1 and the final velocity is below.

v = u + at

v = (164.1662648) + (-9.8)(7.5)
v = 90.6662648 ...

.: Final Velocity is 90.67ms-1


I hope I am the correct.
 

Neha

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ahh sorry i meant with not witch

I see how to do it now, thank you.
 

JulzMighty

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total vertical displacement is the difference between its starting position along the vertical axis and its finishing position, which is zero.

julz
 

twilight1412

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hes finding the initial velocity so he con work it out with

v² - u² = 2as
where s is the height

since the motion is parabolic the initial and final vertical velocities are equal but opposite in direction

btw forbidden i see nothing wrong with your working =)
 

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