need help in projectile motion?

Question

A Projectile reached a maximum height of 50 meters and range from 285 meters. Determine its initial speed and inclination wherein it is fired.

what will I do to get the initial speed and the inclination?

how do I answer it? need help! how do you solve this?

Answer 1

The ratio of four times the maximum height to the range is

tan θ. θ is the inclination of the speed.

tan θ = 4R/ H = 200/285 .

θ is 35°

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The range R = U^2 sin 2 θ /g

U^2 = Rg /sin2 θ = 285 x 9.8 /0.94

U = 54.5m/s

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or

H = [u sin θ]^2 /2g

[u sin θ]^2 = 2gh = 980

[u sin θ] = 31.30

u = 54.5 m/s
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Answer 2

First find the time. Since the projectile falls to the ground from the maximum height, the time to fall is given by the equation:

h = (1/2)gt^2
where g is the acceleration of gravity and h is the maximum height

The total flight time will be twice this since the trajectory is symmetrical. And the horizontal speed (h) will be the horizontal range divided by this time since I am assuming there is no horizontal acceleration.

The vertical speed at impact is then v = gt since the projcetile is falling under the acceleration of gravity

The initial speed is the SQRT(v^2 + h^2) at some angle of inclination to the ground

The inclination is then: tan(angle) = v/h since the initial veritcal velocity will be the opposite of the vertical velocity at impact.

Answer 3

first you must know that projectile motion is a vector quantity, has horizontal and vertical component. The horizontal component is govern only by the initial speed and does not have acceleration. The vertical component is govern by speed and acceleration, the acceleration is the gravity.

Formula for range:
R=Vt cos(angle)
Formula for max. ht
h={[V sin(angle)]^2}/2g

this yields to formula:
R tan(angle) = 4h

there you can solve for angle of inclination, then substitute to solve for initial velocity.

Answer 4

The detrimental sign refers back to the direction of the rigidity, that's downward. by using convention, detrimental velocity and detrimental acceleration are downward. in case you positioned g as effective, then a projectile released upwards might have detrimental velocity, which does no longer make intuitive experience.

Answer 5

H=u²sin²θ/2g=50m---------(1)
R=u²sin2θ/g=285m---------(2)

(1)/(2)

H/R=sin²θ/2sin2θ=50/285

sinθ/4cosθ=10/57
tanθ=40/57

θ=tan^-1(0.7017)=inclination

u(initial velocity) can be found out by substituting the value of
θ in (1) or(2)

Answer 6

how about a hint?

when the projectile reaches 50m height, its velocity in that direction = 0...