Answer to Question #125192 in Physics for qwerty

Question #125192
During a fireworks display, a shell is shot into air at an initial speed of 70 m/s at Angie of 75above the horizontal. The shell ignites once it reaches its highest point. How long did it take to ignite from the time it was launched? answer in seconds
1
Expert's answer
2020-07-06T15:15:10-0400

According to the


https://books.google.com.ua/books?id=QwzwfMr8qncC&pg=PA42&lpg=PA42&dq=a+body+is+thrown+at+some+angle+from+the+ground&source=bl&ots=R_M9bVR11W&sig=ACfU3U2YLUIzaQmA6YBa0BcaKK-Pc4UjgA&hl=ru&sa=X&ved=2ahUKEwjzmtP-p7bqAhVn2aYKHT2DBwYQ6AEwC3oECAkQAQ#v=onepage&q=a%20body%20is%20thrown%20at%20some%20angle%20from%20the%20ground&f=false


the x-coordinate of the shell changes by the following law:


x=v0tcosθx = v_0t\cos\theta

where v0=70m/sv_0 = 70m/s is the initial velocity, tt is time and θ=75°\theta = 75\degree is the launch angle.

The maximum distance that the shell can cover in x-direction is:


xM=v02gsin(2θ)x_M = \dfrac{v_0^2}{g}\sin (2\theta)

Due to the symmetry of the problem, the shell reaches the maximum height when it reaches exactly half of the xMx_M. Thus, substituting the x=0.5xMx = 0.5x_M into the first equation and expressing tt, obtain


v022gsin(2θ)=v0tcosθt=v0sin(2θ)2gcosθ=v02sinθcosθ2gcosθ=v0sinθg\dfrac{v_0^2}{2g}\sin (2\theta) = v_0t\cos\theta\\ t = \dfrac{v_0\sin (2\theta) }{2g\cos\theta} = \dfrac{v_02\sin \theta\cos\theta }{2g\cos\theta} = \dfrac{v_0\sin \theta }{g}

Substituting numerical values, get:


t=70sin75°9.86.9st = \dfrac{70\cdot \sin 75\degree }{9.8} \approx 6.9s

Answer. 6.9 s.


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