You cant answer this?

Question

A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 14.0 m/s, at an angle of theta = 61.0 degrees from the horizontal. The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down (as shown in the Figure). Calculate how long the ball is in the air. ( g = 9.81 m/s2)

Hint: You do not have to separate the upward and downward parts of this motion. You have enough information to find the horizontal and vertical components of initial velocity. You know the vertical displacment and acceleration. Be careful with signs.

Answer 1

The formula to use would be:
d = vt = 1/2 at^2
where d is the vertical displacement, v is the initial; vertical velocity, t is the time, and a is the acceleration (-9.81 m/s^2).
The vertical displacement is 2 meters. The initial upward velocity can be found by multiplying the stated resultant velocity by the sine of 61 degrees.
This equation can give you two possible answers when solving for time...you are looking for the larger of the two (the first answer being when the ball is at 2 meters on the way up, the 2nd being on the way down).

Answer 2

Use the vertical component only.
y = y0+v0*t-.5*9.8*t^2 with
y0 = 0
v0 = 14*sin(61)
Plug these initial values in and use the quadratic formula with
a = .5*9.8
b = -14*sin(61)
c = 2
You will get two values, choose the largest one. The first value is where the ball was at 2 meters going up. Answer: 2.32 s

Answer 3

your right (it would take forever)
but best thing to do is disregard the horizontal motion after you solve for the angle then find out when it would reach 0 mph and then 2 when it would fall 2 meters ... add the two times and you have your answer

Answer 4

Amazing! You're right! I can't! How'd you know?

...and I wouldn't bother if I could...

Answer 5

I can but i wont.

Answer 6

nope.