munkachunka
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Homework Statement
I have made an attempt at the following could someone please check my work.
a mortar shell fired with an angle of elevation of 52 deg has a range of 2.1km, neglecting air resistance, determine.
1, time of flight
2, initial velocity
3, shells range and height after 10 secs
4, the other range at which the height was the same as calculated in part 3 above
Homework Equations
x=u(cos(angle))t
h= u(sin, (angle))t-1/2gt^2
The Attempt at a Solution
a, rearanging x=u(cos52)t to equal x/cos52=ut
1hen x = 2.5km h=0
h=u(sin angle)t-1/2gt^2
substituting 0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2
transposing for t = 23.409
2, x=u(cos52)t therefore u=2100/(cos52)*23.409 = 145.714m/s
3= x=u(cos52)t = 145.712(cos52).10 = 897.093M
height = u(sin52)t-1/2gt^2
= 145.712*(sin52)*10-1/2*9.8*10^2
height = 658.226
4 assuming angle of firing = angle of descent and initial acceleration = rate of deceleration then the other range = 2.1km-897.093 = 1202.907
assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.
your time is greatly appreciated