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f3nr15
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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 ay as -9.8ms-2
Homework Equations
r = uxt
dy = uyt + (ayt2/2)
vy2 = uy2 + 2ah
ux2 + uy2 = u
v = u + at
The Attempt at a Solution
I hope I am the correct
a) r = uxt
ux = r/t
ux = 1200/7.5
ux = 160m
.: Initial Horizontal Velocity is 160m
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-1
c)
ux2 + uy2 = 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 ?