How to find the angle of elevation and launch speed?

Question

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 $u$.

Answer 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