Question

A javelin is thrown at 28.5 m/s from flatground at a 43.2° angle. How much timeis it in the air?(Unit = s)HELP FAST

Answer 1

The initial vertical velocity of the javelin is given by the expression:

`v_(0y)=v_0\sin\theta`

During the flight time, the vertical velocity changes due to the gravitational acceleration. When the javelin touches the ground, the magnitude of its velocity is the same but the direction has changed.

Since the vertical movement of the javelin can be modeled as a uniformly accelerated motion, we can find the time that the javelin is in the air using the equation:

`a=(v_f-v_0)/(t)`

Isolate t from the equation and replace v_f=-v_0:

`t=(v_f-v_0)/(a)=(-v_0-v_0)/(a)=-(2v_0)/(a)`

Replace the expression for v_0 and a=-g:

`t=-(2v_0\sin\theta)/(-g)=(2v_0\sin\theta)/(g)`

Finally, replace v_0=28.5m/s, g=9.81m/s^2 and θ=43.2º:

`t=(2(28.5(m)/(s))\sin(43.2º))/(9.81(m)/(s^2))=3.977...s\approx3.98s`

Therefore, the answer is: 3.98 seconds.