Question

A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal. The target is hit. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the initial veloc- ity?

Answer 1

Answer:

121.7 m/s

Explanation:

Horizontal range, R = 1420 m

Angle of projection, θ = 35°

acceleration due to gravity, g = 9.8 m/s^2

let u be the velocity of projection.

Use the formula for horizontal range

$R=\frac{u^{2}Sin2\theta }{g}$

$1420=\frac{u^{2}Sin70 }{9.8}$

u = 121.7 m/s

Thus, the velocity of projection is 121.7 m/s.

Answer 2

Answer:

$v_{o}=141.51m/s$

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

$x=1420m\\g=-9.8m/s^{2}$

From the equation of x-position we know that

$x=v_{ox}t=v_{o}cos(35)t$

Solving for $v_{o}$

$v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}$ (1)

Now, if we analyze the equation of y-position we got

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0

$0=v_{o}sin(35)t+\frac{1}{2}gt^{2}$

Knowing the equation for $v_{o}$ in (1)

$0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}$

$\frac{1}{2}(9.8)t^{2}=1420tan(35)$

Solving for t

$t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s$

Now, we can solve (1)

$v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s$