How much time is required for the angle to be reduced from 35° at launch to 17°?

Question

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 33 m/s at an angle of 35° above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 35° at launch to 17°?

Answer 1

Assuming no air resistance, the horiz v will be constant at
33*cos(35)

what you want to solve for is when
tan(17)=vy/(33*cos(35))
where vy=33*sin(35)-g*t

Plugging vy in
tan(17)=
(33*sin(35)-g*t)/(33*cos(35))

solve for t
33*(sin(35)-tan(17)*cos(35))/g
1.087 seconds

Note: it will have a -17 degree angle at a later time.

j

Answer 2

because of the fact the object travels upward, its vertical velocity decreases from the preliminary vertical velocity to 0 m/s travels downward, its vertical velocity will develop from 0 m/s to the appropriate vertical velocity. the gap that the object travels upward is comparable to the gap the object travels downward. The time of the upward holiday is comparable to the time for the downward holiday. enable’s use right here equation to be certain the time for the downward holiday. d = vi * t + ½ * a * t^2 For the downward holiday, vi = 0 m/s and a = 9.8 m/s^2. d = ½ * 9.8 * t^2 the gap the object falls is comparable to the optimal top. H = 4.9 * t^2 t = ?(H ÷ 4.9) we can use this time to be certain the appropriate vertical velocity. vf = vi + 9.8 * t, vi = 0 ms/ vf = 9.8 * ?(H ÷ 4.9) that's the object’s very final vertical velocity. So, that's the products preliminary vertical velocity. preliminary vertical velocity = v * sin ? Eq#a million: v * sin ? = 9.8 * ?(H ÷ 4.9) variety = v * cos ? * entire time entire time = 2 * time up = 2 * ?(H ÷ 4.9) variety = v * cos ? * 2 * ?(H ÷ 4.9) Eq#2: v * cos ? = Divide Eq#a million by ability of Eq#2 Tan ? = [9.8 * ?(H ÷ 4.9)] ÷ [variety ÷ 2 * ?(H ÷ 4.9)] you ought to use this equation to be certain the attitude. v = 40 9 m/s and ? = 30? Time up = 40 9 * sin 30 /9.8 = 2.5 s optimal top = ½ * 9.8 * 2.5^2 = 30.625 m variety = 40 9 * cos 30 * 5 = 212.176 enable the kind be 212.176 meters and the optimal top be 30.625 meters Tan ? = [9.8 * 2.5)] ÷ [212.176 ÷ 5)] Tan ? = 24.5 /40 two.4352 ? = 30? This proves that the equation above would be was certain the attitude. you ought to use the attitude to be certain the preliminary velocity. variety = (v^2/9.8) * sin 2 ? 9.8 * variety = v^2 * sin 2 ? 9.8 * variety ÷ sin 2? = v^2 v = ?(9.8 * variety ÷ sin 2?) = ?(9.8 * 212.176 ÷ sin 60 ) ? 40 9 m/s This proves that the equation above is authentic! i desire this has helped you to be attentive to the thank you to remedy this style of situation! in case you decide on my help interior the destiny, make me one among your contacts. Your questions will come directly to my yahoo digital mail tackle!