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.
2006-10-02 15:40:10 · 1 answers · asked by Jay 2 in Science & Mathematics ➔ Mathematics
at time t, the height, in m, is
s = -1/2gt^2 + v0*sin(theta)*t + s0, where theta=61 degrees, s0=0, and v0=14.
To find how long the ball is in the air, you need to find when s=2, so you're solving
-1/2 gt^2 + v0*sin(theta)*t - 2 = 0
or
-4.905t^2 + 12.245t - 2 = 0
Using the quadratic formula, you obtain t=0.176 or t=2.321. The second one is the correct one, because the problem says the ball is caught on the way down (so, at the later time).
2006-10-02 16:10:40 · answer #1 · answered by James L 5 · 0⤊ 0⤋