A projectile is launched from ground level at 38.2 m/s at an angle of 38.7 ° above horizontal. Max Height?
A projectile is launched from ground level at 38.2 m/s at an angle of 38.7 ° above horizontal. Use the launch point as the origin of your coordinate system.
How far downrange (the horizontal distance from the origin) was the projectile when it reached the highest point in its flight?
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The projectile has a vertical component of velocity given by:
v sin(theta) = 38.2 x sin (38.7)
at maximum height the velocity is zero. We can call this the final velocity.
v^2 = u^2 - 2as
0 = [38.2 x sin (38.7)]^2 - 2 x 9.81 x s
The negative sign is because gravity is decelerating the object. You can solve this expression for s.
OK, how long does it take to get to maximum height.
v = u + gt
0 = 38.2 x sin (38.7) - 9.81 x t
Solve this for time. Now, the projectile has a horizontal component of velocity given by
v cos (theta) = 38.2 x cos(38.7)
Multiply this by the time to reach maximum height, and you'll get the down range distance when the projectile is at max height. The flight is called parabolic, and is symmetric about the apex (maximum height). Does this help? Remember, projectiles are easy. You just need to use the horizontal and vertical components. The vertical component changes because gravity acts on the object. The horizontal component does not, as no force acts on it. This only works when there is no air resistance.
First, rockets are not projectiles. Projectiles won't be self propelled; they're projected by technique of different mindset. for this reason the term, projectile. 2d, via fact the projectile is acted on Most worthy by technique of the vertical force of gravity (discounting drag), the time of flight T is set thoroughly by its vertical circulation over the years. we are able to write that as: y(t) = h + Uyt - a million/2 of gt^2; the situation h is the launch top above the influence top, Uy = U sin(theta) is the vertical preliminary %. = 30 sin(50) and U = 30 mps, g is g, and t = T is the time of flight you're searching for. you're able to desire to understand the physics of the y(t) equation. the 1st term is the initiating top at t = 0. the 2d term is the top the projectile might pass if there have been no gravity. The final term is how gravity impacts the vertical flight of the projectile (e.G., pulling it back down). via fact the effect is believed to be floor degree, we set y(T) = 0 = h + (U sin(theta))T - a million/2 of gT^2 and then treatment for T given the theory h = 0 additionally. So we've U sin(theta) = a million/2 gT and T = 2U sin(theta)/g = 2*30*sin(50)/9.80 one = 4.sixty 9 sec ANS.
just produce your ballistics equation, then differentiate. When the grad = 0 that gives you the distance downstream at maximum height