Answer to Question #107167 in Mechanics | Relativity for elvis barry

Question

A canon is launched with an initial velocity of 25m/s and at an angle of 53° to the horizontal on a platform 60m high. Find:

I) the maximum height

II) range of the canon

III) The time for the canon to land

Answer 1

Given:

$v_0=25\:\rm m/s;$

$\theta=53^{\circ};$

$y_0=60\:\rm m.$

(i) The equations of motion of the canon

$y(t)=y_0+v_{0}\sin\theta t-\frac{gt^2}{2}=60+20t-5t^2;\\ x(t)=x_0+v_{0}\cos\theta t=15t.$

Let's find the maximum value of $y(t)$:

$y'(t)=20-10t, \longrightarrow t=2\:\rm s.$ $y_{max}=y(2)=60+20\times 2-5\times (2)^2=80\:\rm m.$

(iii)

$y(t)=60+20t-5t^2=0, \longrightarrow t=6\:\rm s.$

(ii) Range

$R=x(6)=15\times 6=90\:\rm m.$