A particle is projected from a point on level ground with a speed of $u$ meters per second and an angle of elevation $\theta$. The maximum height reached by the particle is $42$ meters above the ground and the particle hits the ground $196$ meters from its point of projection. Find the values of $\theta$ and $u$.
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I would approach this as follows (if lacking formulas derived earlier). 1) calculate the time $T_1$ it takes an object to fall 42 meters, starting from rest. You should then know that the total flight time is $T=2T_1$. 2) You also know that the vertical component of the initial speed is $gT_1$. 3) If you know the flight time and the distance travelled, you can get the horizontal component of the initial speed easily, because that is constant (assuming no air drag here). 4) You know the $x$ and $y$-components of the initial speed, so basic trig and Pythagoras give you $u$ and $\theta$. – Jyrki Lahtonen Sep 10 '13 at 20:18
1 Answers
1
let "a" be angle as its easy to type for me
Horizontal
u(x)= u cosa
a(x)=0
s(x)= 196 m
Vertical u(y)= u sina a(y)= -9.8 s(Y)= 0 M
S= u*t + 0.5 a t^2
first vertically at max height s(y)=42 and time be t
42= u sina* t -4.9t^2
u sina *t =42+ 4.9t^2 .....(1)
now basically the time is doubled from the maximum height as it's coming to ground so we use 2t
first Horizontally
196= u cosa *2t ........(2)
now vertically
0= u sina *2t -4.9 (2t)^2
u sina 2t = 4.9 4 t^2 .... (3)
from equations 1 and 3
2* (42+4.9 t^2) = 4.9* 4 t^2
t= 8.57 sec
now sub in 2 u cosa * 2*8.57 = 196
sub in 3 u sina* 2*8.57 = 4.9 4 8.57^2
divide each other Tana= 4.9 4 8.57^2/ 196 = 7.35
a= 82.2 degrees
sub back in 2 or 3 to get u= 84.8 m/s
Mr. Math
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