Question

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

Answer 1

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$. In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$.

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$.

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$.

Answer 2

Answer:

4.21 s

Explanation:

Vertical component of velocity = 36 sin 35 = 20.649 m/s

Vertical position is given by

yf  = y0 +          vo t         - 1/2 at^2                yf = yo = 0  (ground level)

0  = 0  +   20.649 m/s * t - 1/2(9.81)t^2

     t ( 20.649 - 4.905 t) = 0     show t = 0   and   4.21  s

              the t = 0 is launch      4.21 seconds is landing