Answer to Question #168290 in Calculus for Shein

Question

A ball is thrown vertically upward from the ground with an initial velocity of 64 ft/sec. If the positive direction of the distance from the starting point is up,

the equation of the motion is s =-16t²+64t. Let t seconds be the time that has elapsed since the ball

was thrown and s feet be the distance of the ball from the starting point at t seconds.

A. Find the instantaneous velocity of the ball at the end of 1 sec. Is the ball rising or talling at the end of 1 sec?

B. Find the instantaneous velocity of the ball at the end of 3 sec. Is the ball rising or falling at the end of 3 sec?

C. How many seconds does it take the ball to reach its highest point?

D. How high will the ball go?

E. Find the speed of the ball at the end of 1 sec and 3 sec.

F. How many seconds does it take the ball to reach the ground?

G. Find the instantaneous velocity of the ball when it reaches the ground.

Answer 1

Given $s = -16t^2+64t$ :

A.

$$v = \frac{ds}{dt} =-32t+64$$

at the end of 1sec,

$$v= -32(1) +64\\ v=32ft/sec$$

Since v is positive, then the ball is rising after 1 sec

B. at t=3

$$v= -32(3) +64\\ v=-96 + 64\\ v= -32ft/sec$$

After t=3, the ball is falling

C. At highest point, v=0,

$$\therefore -32t+64 = 0\\ 32t = 64\\ t=2 secs$$

at t=2 secs the ball reach its highest point

D. Since the ball reach it's highest point at t=2sec, then the maximum height of the ball is:

$$s = -16t^2+64t\\ s = -16(2)^2+64(2)\\ s= -16(4)+128\\ s= -64+128\\ s=64ft$$

E. The speed at the end of 1sec and 3 sec

$$Speed_{1sec} = \frac{ds}{dt}\\$$

(a) at the end of 1 sec

$$\frac{ds}{dt} = -32t+64\\ = -32(1)^2+64(1)\\ = -32+64\\ = 32ft/sec$$

(b) at the end of t=3secs

the speed at the end of 3secs

$$Speed_{3secs} = \Big|\frac{ds}{dt} \Big|\\ = \big| -32t+64\big|\\ = \big| -32(3)+64\Big|\\ = \big| -96+64\big|\\ = |-32|\\ = 32ft/sec$$

F. at the ground level, s= 0,

$$s = -16t^2+64t\\ -16t^2+64t =0\\ -16t(t-4) = 0 \implies 16t =0 \quad or \quad t-4=0\\ \implies t=0 \quad and \quad t=4$$

 Since the ball is returning to the ground, then at t=4secs, the ball hits the ground

G. The instantaneous velocity of the ball when it reaches the ground.

Since: $v = -32t+64$ and the ball reach the ground at t=4 secs, then:

$$v = -32(4)+64\\ v= -128+64\\ v= -64ft/sec$$