Asked by Anonymous
A ball is thrown at an angle of 45 degrees to the ground. If the ball lands 90m away, what is the initial speed of the ball? I know you use x(t)=(v*cos(alpha))t and y(t)=(v*sin(alpha))t-g/2t^2 but I'm still really confused about what to do. Please show all work. Thank you.
Answers
Answered by
MathMate
The question is missing information.
By matching different initial velocities of the ball, it can land at 90m away at virtually any angle α.
I will solve for the initial velocity in terms of the angle α, but also give a solution for the MINIMUM initial velocity for the ball to land 90m away.
The key is to recognize that
x(t)=90m=D, and
y(t)=0
(i.e. the ball lands D=90m away).
Using the second formula, we have
y(t)=(v*sin(alpha))t-(g/2)t^2, or
(v*sin(alpha))t=(g/2)t^2
Solve for t after cancelling t on each side:
t=2vsin(α)/g
Substitute in the first equation to solve for v(cos(α):
D=vcos(α)*(2vsin(α)/g
Substitute
sin(2α) = 2sin(α)cos(&alpha)
D=v²sin(2α)/g
or
v²=Dg/sin(2α)
(The following is the only place where calculus comes in)
The minimum velocity that can give the distance D is obtained when the ball is thrown at 45° with the horizontal, in which case α=45°, or when 2α=90°.
In this case the (minimum) initial velocity is
v=sqrt(90*9.81)=29.71m/s
By matching different initial velocities of the ball, it can land at 90m away at virtually any angle α.
I will solve for the initial velocity in terms of the angle α, but also give a solution for the MINIMUM initial velocity for the ball to land 90m away.
The key is to recognize that
x(t)=90m=D, and
y(t)=0
(i.e. the ball lands D=90m away).
Using the second formula, we have
y(t)=(v*sin(alpha))t-(g/2)t^2, or
(v*sin(alpha))t=(g/2)t^2
Solve for t after cancelling t on each side:
t=2vsin(α)/g
Substitute in the first equation to solve for v(cos(α):
D=vcos(α)*(2vsin(α)/g
Substitute
sin(2α) = 2sin(α)cos(&alpha)
D=v²sin(2α)/g
or
v²=Dg/sin(2α)
(The following is the only place where calculus comes in)
The minimum velocity that can give the distance D is obtained when the ball is thrown at 45° with the horizontal, in which case α=45°, or when 2α=90°.
In this case the (minimum) initial velocity is
v=sqrt(90*9.81)=29.71m/s
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