What was the ball's initial speed?

Question

A spring gun at ground level fires a golf ball at an angle of 45 degrees. The ball lands 10 m away.

a) What was the ball's initial speed?

b) For the same initial speed, find the two firing angles that make the range 6 m.

Recall that the Ideal Projectile Motion Equation is

r=(vo*cos(theta))ti+((vo*sin(theta)t-1/2*g*t^2)j.

Answer: a) vo=sqrt(10g)

b) theta=1/2*arcsin(3/5),

theta=pi-1/2*arcsin(3/5).

And is arcsin the same thing as sin^-1?

Answer 1

a)

Horizontal direction: v_o cos45 t = 10 ......(1)

Vertical direction: v_o sin45 t - (1/2)g t² = 0 ......(2)

Since sin45 = cos45,

10 = (1/2)gt² ==> t = √(20/g)

Plug in (1),

v_o√(10/g) = 10

Solve for v_o,

v_o = √(10g)

 

b)

Horizontal direction: v_o cosθ t = 6 ......(1)

Vertical direction: v_o sinθ t - (1/2)g t² = 0 =>Since t ≠ 0,  t = 2v_o sinθ/g......(2)

Plug in (2) in (1),

v_o²sin2θ/g = 6 => 10sin2θ = 6, since v_o² = 10g.

sin2θ = 3/5

θ = (1/2)arcsin(3/5)

or

2θ = pi - arcsin(3/5)

θ = pi/2 - (1/2)arcsin(3/5)

Yes, arcsin(x) = sin^-1(x)

 

 

Answer 2

For the vertical component (v _y) of the initial speed(v ₀) we can write:         v_y = v₀ sin θ, where θ = 45⁰ = √2/2.

For the horizontal component we have: v _x = v₀ cos θ. We have following equations for the vertical and horizontal motions:

                              v _y = gt, g = 9.8 m/sec ² .  Or v₀ sin θ = gt. Thus,  t = v₀ sin θ/g

In the neantime          L = 2v_xt = 2v₀t cos θ..

Substitute t into last expresion. We will obtailn:    L = 2V₀² sinθ cosθ/g. Using trigonometric identity

                      2 sinθ cosθ = sin2θ = sin 90 ⁰ = 1  we then can wriite

                    v₀² = gL    and therefore     v ₀ = sqrt(9.8 x10) = 9.9 m/sec

For two other angles when L = 6m you can write

                              sin 2θ = gL/v₀² = 9.8x6/98 = 0.6

Two angles are  18.43 ⁰ and 71.57 ⁰ . They are complimentary angles. 

Answer 3

Hey Sun!  We can use some physics reasoning here. Let g= 10 m/s^2 ...

a) At 45 deg, v up = v right. At 10 m/s up, ball is in the air 2 sec -- 1 sec up as v up slows from 10=>0 and then another second down. 10^ means 10>, and at 2-sec airtime would cover 20 m => too far ...

Try v up = 5 m/s, ball airborne 0.5 sec up and 0.5 sec down. One sec airtime covers 5m => too short.  Let's try v up 7 m/s, ball airborne 1.4 sec, making range of 9.8m nearly to spec ==> v= 10 m/s /45.

b) For same initial speed, you could try almost horizontal 3^9.5> where sin is 0.3 rads or /18.  Of course, the almost straight up 9.5^3> at /72 would give the same approx. range of 5.7 m or so.  Regards :)