1. 0.5g*t^2 = 2010 m.
4.9t^2 = 2010.
t = 20.3 s. = Fall time.
D = Xo*t. = 193m/s * 20.3s = 3909 m.
2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s.
3. Vo*t + 0.5g*t^2 = 2010 m.
58*t + 4.9*t^2 = 2010.
4.9t^2 + 58t - 2010 = 0.
Use Quadratic Formula.
t = 15.2 s. = Fall time.
D = 193m/s * 15.2s = 2934 m.
The initial velocity is 
Explanation:
The motion of the ball is a projectile motion, therefore it consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
In this problem, we just need to analyze the horizontal motion: the horizontal velocity is constant, therefore the horizontal distance travelled is given by

where
is the horizontal velocity
t is the time of flight
Here we have:
t = 0.45 s
x = 3.0 m
And so solving for
, we find

And since the ball was initially projected horizontally, this is also the initial velocity.
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
Newton meter
Torque wrench
Or Just a plain Scale
The best and most correct answer to the question is :
Alpha particles
Beta particles
Gamma rays
Cosmic radiation
Neutrons
Hope this helped you :)