Problem 1C-07*: Two-dimensional free throw problem without air resistance
A general free throw problem is a two-dimensional problem. If a ball is thrown, it travels within a vertical plane, so the motion can be described by a horizontal x-axis and a vertical y-axis defined within that vertical plane. The vertical position of the ball is denoted as y(t) and the horizontal position denoted as x(t). As discussed in Problems 1A-06, 1A-07 and 1A-12, y(t) and x(t) are described by functions of t as
where g is a positive constant the meaning of that will become clear later.
(a) If the ball is thrown at t = 0 with a speed of v0 and an angle θ from a height h0, where will the ball land?
(b) How high will the ball reach?
(c) If the thrower wants to maximize the horizontal landing distance, at what angle should the ball be thrown?
Scroll down for solution
Solution:
(a) Eq.(1) leads to
From the initial condition vy(0) = v0sinθ and vx(0) = v0cosθ, we have B = v0sinθ and D = v0cosθ. From Eqs.(1) we have
To calculate time T that the ball lands, we need to solve y(T) = 0. From the first equation of Eqs.(3), we have
The minus sign of the above equation is not acceptable since it will make T ‹ 0. So we must have
Since vx(t) does not depend on time t, the horizontal landing
distance, L, is L = vx(t)·T. Substituting the last equation
of Eqs.(3) and T of Eq.(4) into this relation, we get
(b) Suppose that the ball reaches the maximum height at time t = tmax. We must have vy(tmax) = 0. The third equation of Eqs.(3) then says that
Substituting tmax into y(t) of Eqs.(3), we get the maximum height, hmax as
(c) Put z ≡ sinθ. Eq.(5) can be rewritten as
Let us define
Differentiating L by z, we get
L increases as z increases from zero. Thus the maximum L is reached when dL/dz = 0. Equating dL/dz of Eqs.(6) to zero, we get
This last equation can be simplified to become
is the initial angle that makes the ball land farthest.
| <-Previous page | Table of contents | Next page-> |