Physics, 31.07.2019 15:30 raquelqueengucci25

Aprojectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. using g = 9.8 m/s2 , the distance from launching to landing points is:

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Answer from: QueenNerdy889

The distance from launching to landing point (range) is 122.47 m

To calculate the range of the projectile, first, we need to find the angle of projection using the formula below.

Formula:

∅ = tan⁻¹(Vy/Vx) Equation 1

where:

∅ = Angle of projection.

Given:

Vy = 20 m/sVx = 30 m/s

Substitute these values into equation 1

∅ = tan⁻¹(20/30)∅ = tan⁻¹(0.667)∅ = 33.7°

Finally, we use the formula below to calculate the range of the projectile.

R = V²sin2∅/g Equation 2

Where:

R = Range of the projectile.V = Resultant velocity∅ = Angle of projectiong = acceleration due to gravity.

From the question,

Given:

V = √[(20²)+(30²)] = √(400+900) = √1300 m/s∅ = 33.7°g = 9.8 m/s².

Substitute these values into equation 2

R = (√1300)²sin(2×33.7)/9.8R = 1300(sin67.4)/9.8R = 1200.17/9.8R = 122.47 m.

Hence, The distance from launching to landing point (range) is 122.47 m.

Learn more about the range here: link

Answer from: CrownedQueen

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s

, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s

and acceleration g=9.81 m/s^2

$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0

Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
t=0

is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m

S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m

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