Chapter 4.2, Problem 43PPSE

Solution

To compare: The heights of two projectiles: which projectile higher and how much higher.

The projectile- 1 goes higher at height 64 ft in 2 seconds.

Given data: The height h of projectile-1 fired straight up with initial velocity $\text{64}\text{ft/}{\text{s}}^{\text{2}}$ is

  $h=64t-16t^2$ , the height h in ft and time t in seconds. The data for the height of projectile-2 is

Time ( t ) in seconds	0.5	1.0	1.5	2.0
Height ( h ) in ft	20	32	36	32

Formula used:

The height of the projectile at straight up height h is given by $dh/dt$ .

Calculation:

The height h of projectile-1 fired straight up with initial velocity $\text{64}\text{ft/}{\text{s}}^{\text{2}}$ at time t is

  $h\left(t\right)=64t-16t^2$ … (1)

Differentiating with respect to‘ t’

  $\begin{array}{c}\frac{dh}{dt}=64-16\left(2t\right)\\ =64-32t\end{array}$

Thus, the velocity $v\left(t\right)$ of the projectile at any time t is

  $v\left(t\right)=64-32t$ … (2)

The projectile will continue going up until its velocity becomes zero. At that time the height will be maximum.

Substitute 0 for $v\left(t\right)$ in equation (2),

  $\begin{array}{c}0=64-32t\\ 32t=64\\ t=\frac{64}{32}\\ t=2\end{array}$

Thus after 2 second the projectile-1 will be at maximum height. Therefore, the maximum height of the projectile-1 is

  $\begin{array}{c}h\left(2\right)=64\left(2\right)-16{\left(2\right)}^2\\ =64\left(2\right)-64\\ =64\end{array}$

Thus, the maximum height attain by the projectile-1 is 64 ft when $t=\text{2}\text{.0}\text{s}$

From table the maximum height of the projectile-2 is 36 ft when $t=\text{1}\text{.5}\text{s}$ .

Therefore, the projectile- 1 goes higher at height 64 ft in 2 seconds.