need projectile motion help, dealing with change in angle, equation, explanation and solution requested?

Question

In the javelin throw at a track and field event, the javelin is launched at a speed of 31 m/s at an angle of 36° 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 36° at launch to 19°?

Answer 1

First lets break up the initial velocity into horizontal and vertical components:
v(horz) = 31cos36 = 25.0795m/s
v(zero vert) = 31sin36 = 18.2213m/s

At 19 degrees we want:
tan19 = v(vert 19)/v(horz)
v(vert 19) = (tan19)(v(horz)) = tan19(25.0795) = 8.63556m/s

Vertical:
v(vert 19) = v(zero) + at
8.63556 = 18.2213 + (-9.8t)
-9.58574 = -9.8t
t = 0.978sec

Answer 2

entire velocity V = sqrt(Vx^2 + Vy^2); the place Vx is the X factor (horizontal) and Vy is the Y factor (vertical). Vx is in many circumstances seen consistent by using fact the subject question is discounting all forces on the projectile that act in the horizontal (X) direction. as an occasion, in a real case, there may be drag forces by using fact the projectile cuts via the air. yet HS and early college physics questions many times forget approximately approximately them to make the respond greater undemanding to return via. And to a element, even in the genuine case, the horizontal forces are many times no longer significant; so as that they may be passed over if relatively some precision isn't required in the respond. Vy on the different hand is brought about via the only significant tension appearing on a projectile after it leaves the muzzle...gravity. And as you probable comprehend, the better an merchandise falls from, the quicker would be its velocity while the object strikes the floor. And, that is significant, a projectile falls to the floor decrease than the tension of gravity purely like all different merchandise falling from some top. And the attitude of the shot wrt the horizontal is a significant factor in determining purely how extreme above floor the projectile will circulate. that is common to tutor.... assume Vy = V sin(theta); the place theta is that capturing attitude. we are in a position to tutor that Vy^2 = 2gH; the place H is the max top the projectile will circulate above the muzzle. consequently, H = Vy^2/2g = (V sin(theta))^2/2g and there you have it, H relies upon partly on sin(theta). as an occasion, if the gun is pointing point with the floor, theta = 0 and H = 0 for this reason. And if the gun is pointing at present up with theta = ninety deg, H = V^2/2g.

Answer 3

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