Question

A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can be ignored, how high does it go? The acceleration due to gravity is 32 ft per second squared.

Would you please solve this question using CALCULUS, NOT physics equations?

Answer 1

Answer:

Maximum height of the ball, h(t) = 27.56 m

Explanation:

It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.      

The height of the ball as a function of time t is given by :

$h(t)=h_o+v_ot-16t^2$

h₀ is initial height, h₀ = 0

So, $h(t)=42t-16t^2$ .........(1)

For maximum/minimum height,  $\dfrac{dh(t)}{dt}=0$

$42-32t=0$...(2)

t = 1.31 s

Differentiating equation (2) wrt t

h''(t) = -32 < 0

So, at t = 1.31 seconds we will get the maximum height.

Put the value of t in equation (1)

$h(t)=42\times 1.31-16\times (1.31)^2$

h(t) = 27.56 m

Hence, this is the required solution.

Answer 2

The maximum height reached by the ball is 27.56 ft.

The given parameters;

• initial velocity of the ball, u = 42 ft/s
• acceleration of the ball, a = 32 ft/s²

The maximum height reached by the ball is calculated as follows;

v² = u²- 2gh

where;

v is the velocity of the ball at maximum height = 0

0 = u² - 2gh

2gh = u²

$h = \frac{u^2}{2g} \\\\h = \frac{(42)^2 }{2\times 32} \\\\h = 27.56 \ ft$

Thus, the maximum height reached by the ball is 27.56 ft.

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