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

Question #107167
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
1
2020-04-03T13:23:44-0400

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;\\\\\nx(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."

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