a particle is projected at an angle tita to the horizontal with an initial velocity U.if the maximum horizontal distance travelled is 20m and the greatest height reached is 10m.find U and Tita

Answer 1

the horizontal distance is

x(t) = u cosθ t
The time taken to go 20m is thus
t = 20/(u cosθ)

The trajectory is
y(t) = u sinθ t - g/2 t^2
We know that y=0 at 0 and 20/(u cosθ), so the max height is reached at 10/(u cosθ). So,

u sinθ * 10/(u cosθ) - g/2 (100/(u^2 cos^2 θ) = 10

y = tanθ x - g/(2 (u cosθ)^2) x^2

We know that y(20) = 0 and y(10) = 10. So, approximating g as 5 m/s^2,

20tanθ - 500/(u cosθ)^2 = 0
10tanθ - 125/(u cosθ)^2 = 10

u = 5/2 √10
tanθ = (√5-1)/2

Answer 2

Initial vertical speed Vi = U sin theta
constant horizontal speed = u = U cos theta

horizontal problem
20 = U cos theta * T where T is time in air

vertical problem
v = Vi - g t
v = 0 at the top
so
t at top = Vi/g
g = 9.8 m/s^2 on earth
so
t at top = Vi/9.8

now at the top at 10 meters
h = 0 + Vi t -4.9 t^2
10 = Vi t - 4.9 t^2
10 = Vi^2/9.8 - 4.9 Vi^2/96
10 = Vi^2 (.051)
Vi = 14 m/s
t at top = 14/9.8 = 1.43 s
total time in air = 2t = 4.29 s
20 = u (4.29)
so u = 4.67 m/s
so
U = sqrt (4.67^2+14^2) = 14.8 m/s
and
tan theta = 14/4.67
theta = 71.6 degrees

Answer 3

t at top = 14/9.8 = 1.43 s
total time in air = 2t = 2.86 s
20 = u (2.86)
so u = 6.99 m/s
so
U = sqrt (6.99^2+14^2) = 15.64 m/s
and
tan theta = 14/6.99
theta = 63.5 degrees