Hi there!
Since the ball is thrown with an initial HORIZONTAL velocity, we can treat this as a free-fall situation since the horizontal motion does NOT impact the ball's vertical motion.
We can use the derived kinematic equation:
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
Plug in the given values:
[tex]t = \sqrt{\frac{2(150)}{9.8}} = \large\boxed{5.53 s}}[/tex]
The angle of projection is 0° because there is no vertical component to the velocity as the initial velocity is purely horizontal.
Range:
We can use the equation:
dₓ = vₓt
displacement (x direction) = velocity (x direction) · time
Used the solved-for time and given velocity:
dₓ = 100 · 5.53 = 553 m
