Question

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The time units are seconds and the distance units are feet. The distance between the location of the pitcher and homeplate (where the batter stands) is 60.5 feet. Give EXACT answers, unless instructed otherwise. (a) Calculate the horizontal velocity of the baseball at time t; this is the function x'(t)= 142 Correct: Your answer is correct. ft/sec. (b) What is the horizontal velocity of the baseball when it passes over homeplate? 142 Correct: Your answer is correct. ft/sec (c) What is the vertical velocity of the baseball at time t; this is the function y'(t)= $$−32t+5 Correct: Your answer is correct. -32t +5 ft/sec. (d) Recall that the speed of the baseball at time t is s(t)=√ [x '(t)]2 + [y ' (t)]2 ft/sec. What is the speed of the baseball (in mph) when it passes over homeplate? $$1 Incorrect: Your answer is incorrect. (sqrt(142*142 +((-1936/142) + 5)**2)*(360/528)) mph. (e) At what time does the baseball hit the ground, assuming the batter and catcher miss the ball? $$1.1 Incorrect: Your answer is incorrect. (5+sqrt(5**2 + 320))/32 sec. (f) What is the magnitude of the angle at which the baseball hits the ground? 0.12 Incorrect: Your answer is incorrect. rad. (This is the absolute value of the angle between the tangential line to the path of the ball and the ground. Give your answer in radians to three decima

Answer 1

Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$.

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$.

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$