Question

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

Answer 1

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:

$$x = v_0t\cos\theta$$

where $v_0 = 70m/s$ is the initial velocity, $t$ is time and $\theta = 75\degree$ is the launch angle.

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

$$x_M = \dfrac{v_0^2}{g}\sin (2\theta)$$

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

$$\dfrac{v_0^2}{2g}\sin (2\theta) = v_0t\cos\theta\\ t = \dfrac{v_0\sin (2\theta) }{2g\cos\theta} = \dfrac{v_02\sin \theta\cos\theta }{2g\cos\theta} = \dfrac{v_0\sin \theta }{g}$$

Substituting numerical values, get:

$$t = \dfrac{70\cdot \sin 75\degree }{9.8} \approx 6.9s$$

Answer. 6.9 s.