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At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at

Agata [3.3K]

We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.

8 0

3 years ago

Read 2 more answers

How to determine the amount of resistance when two or more resistors are in parallel.

Elena L [17]

Answer: 1/R = 1/R1 + 1/R2+ ...+ 1/Rn

R is resistance of system in which there are resistors R1, R2 , ... Rn parallel.

3 0

3 years ago

A car is traveling at 16 m/s^2

Leya [2.2K]

Really? wow that's pretty cool

4 0

3 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i

3 0

3 years ago

Read 2 more answers

Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0

3 years ago