Question

A softball is hit over a third baseman's head with some speed v0 at an angle θ above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V=7.00m/s. He catches the ball t=2.00s later at the same height at which it left the bat. The third baseman was originally standing L=18.0m from the location at which the ball was hit.

A) Find v0. Use g=9.81m/s2 for the magnitude of the acceleration due to gravity.

Express the initial speed numerically in units of meters per second to three significant figures.

B) Find the angle θ in degrees.

Express your answer numerically in degrees to three significant figures.

Answer 1

Answer:

A) v0 = 18.8 m/s

B) θ = 31.5°

Explanation:

On the x-axis:

$X_{ball} = X_{player}$

$v0*cos\theta*t=L+V*t$

where

t = 2s;  L = 18m;  V = 7m/s

$v0*cos\theta*2=32$

$v0=16/cos\theta$

On the y-axis for the ball:

$\Delta Y=v0*sin\theta*t-1/2*g*t^2$

$0 = v0*sin\theta*2-19.62$

Replacing v0:

$0 = tan\theta*2*16-19.62$

$\theta=atan(19.62/32)=31.5\°$

Now, the speed v0 was:

v0 = 18.8m/s

Answer 2

Solution is:

• V₀ = 16.735 m/s

• θ = 17.0403 °  

The events  "hitting the ball"  and "start running" ( the third baseman ) are simultaneous.

• Calculating the distance x of the fly

the distance the third baseman run is:

x₁  = v × t          v = 7 m/s    and t = 2 sec

Then    x₁ =  7 × 2  = 14 mts

The original distance between the bateman and the third baseman was L = 18 m, the ball fly for 2 seconds and went through the distance

x = 18 + x₁   = 32 m

Total distance x of the fly   32 m

The time of the fly is: 2 sec

then

x = v₀ₓ × t            32  = v₀ₓ × 2          v₀ₓ = 16 m/s  

In parabolic shooting  v₀ₓ = V₀ × cosθ      is a constant speed

• Calculating the shooting angle

On the other hand if the fly last 2 sec., by symmetry the maximum height was at 1 sec after hitting the ball. At y maximum  Vy = 0

Vy = v₀y - (1/2)× g×t

Vy = 0   = v₀y - g/2                v₀y = g/2       v₀y = 4.905 m/sec

v₀y = V₀ × sinθ         (1)

v₀ₓ  = V₀ × cosθ       (2)

Dividing equation (1) over equation (2)

(4.905/16) = tan θ = 0.3065

Then  θ = arctan ( 0.3065)

θ = 17.0403 °       or   θ = 17°2´25.088"

cosθ = 0.9561      ( from tables)

• Calculating initial speed

and finally    v₀ₓ = V₀ × cosθ

Then     V₀  = v₀ₓ / cosθ

V₀  =  16 / 0.9561 m/s

V₀  = 16.735 m/s