Question

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

0 meters). Calculate the time the cannonball spent in the air. Assume no air resistance and
that the cannonball is launched on Earth (g = -9.81 m/s²).

Answer 1

4.21 s

Step-by-step 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

Answer 2

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

Step-by-step 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 &= (\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}`.