A projectile is shot and lands 15 m away from the starting point.

A projectile is shot and lands 15 m away from the starting point. If the projectile is timed and takes 1.2 seconds. What is the initial velocity and the angle that the projectile is shot at?

(15/1.2) = horizontal component of 12.5m/sec.
The projectile will rise for (1.2/2) = 0.6 sec.
It will gain a height of 1/2 (t^2 x g) = 1.764 metres.
Its initial vertical velocity component = sqrt. (2gh) = 5.88m/sec.
Initial velocity = sqrt. (12.5^2 + 5.88^2) = 13.814m/sec., and initial angle = arctan (5.88/12.5) = 2.693 degrees above horizontal.

U = initial velocity (vertical component =Uy , horizontal = Ux)

1.2/2 = Uy/g ==> Uy = 0.6*9.8 = 5.88 m/s
15 = Ux * 1.2 ==> Ux = 12.5 m/s
U² = Ux² + Uy² = 12.5² + 5.88² = 190.8244
U = 13.814 m/s ............answer for initial velocity

tanA = Uy /Ux = 5.88 /12.5
A = 25.2° .......answer for angle
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Dunno

Vo= 12.5m/s , i forget how to find the angle