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17. A projectile is fired aran angle of 35^circ relative to the horizontal (upwards) with a speed of 20m/s (linore air resistance,Calculate the a. horizontal componen of the projectile's initial velocity, [16m/s] b. vertical component of the projectile's initial velocity. [11m/s] c. time the projectile is in the air. [2.] I d. horizontal distance the projectile travels (Le, it range), [3B m]

Question

17. A projectile is fired aran angle of 35^circ relative to the horizontal (upwards) with a speed of 20m/s (linore air resistance,Calculate the a. horizontal componen of the projectile's initial velocity, [16m/s] b. vertical component of the projectile's initial velocity. [11m/s] c. time the projectile is in the air. [2.] I d. horizontal distance the projectile travels (Le, it range), [3B m]

17. A projectile is fired aran angle of 35^circ  relative to the horizontal (upwards) with a speed of
20m/s (linore air resistance,Calculate the
a. horizontal componen of the projectile's initial velocity, [16m/s]
b. vertical component of the projectile's initial velocity. [11m/s]
c. time the projectile is in the air. [2.] I
d. horizontal distance the projectile travels (Le, it range), [3B m]

Solution

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DavidVeteran · Tutor for 11 years

Answer

Here's the solution:<br /><br />**a. Horizontal component of initial velocity:**<br /><br />* The horizontal component (vₓ) is found using the cosine of the launch angle: vₓ = v * cos(θ)<br />* vₓ = 20 m/s * cos(35°) <br />* vₓ ≈ 16.38 m/s ≈ **16 m/s** (rounded to two significant figures as per the provided answer)<br /><br />**b. Vertical component of initial velocity:**<br /><br />* The vertical component (vᵧ) is found using the sine of the launch angle: vᵧ = v * sin(θ)<br />* vᵧ = 20 m/s * sin(35°)<br />* vᵧ ≈ 11.47 m/s ≈ **11 m/s** (rounded to two significant figures)<br /><br />**c. Time the projectile is in the air:**<br /><br />* The time of flight (t) can be found using the vertical component of the velocity and the acceleration due to gravity (g = 9.8 m/s²). The time it takes to reach the highest point is equal to the time it takes to fall back down. We can use the following equation: v_f = v_i - gt, where v_f is the final vertical velocity (0 m/s at the highest point), and v_i is the initial vertical velocity.<br />* 0 = vᵧ - gt<br />* t = vᵧ / g<br />* t = 11.47 m/s / 9.8 m/s²<br />* t ≈ 1.17 s This is the time to reach the highest point. The total time in the air is twice this:<br />* Total time = 2 * t ≈ 2 * 1.17 s ≈ 2.34 s ≈ **2.3 s** (The provided answer of 2.1s seems to have used a rounded value of 11 m/s for the vertical velocity. Using the more precise 11.47 m/s gives a more accurate answer of 2.3s)<br /><br />**d. Horizontal distance the projectile travels (Range):**<br /><br />* The range (R) is the horizontal distance traveled and is calculated using the horizontal velocity and the total time of flight: R = vₓ * t<br />* R ≈ 16.38 m/s * 2.34 s<br />* R ≈ 38.3 m ≈ **38 m** (The provided answer of 38m is consistent with using the slightly less precise values from earlier calculations. If we use the more precise values, we get a range closer to 38m).<br />
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