<p>A football is kicked off the flat ground at 30.0m/s^2  at an angle of 30
ID: 2053024 • Letter: #
Question
<p>A football is kicked off the flat ground at 30.0m/s^2  at an angle of 30 degrees relative to the ground.</p><p>
<div class="partText">a.) Determine the total time it is in the air.</div>
<div class="partText">
<div class="partText">b.)Find the angle of its velocity with respect to the ground after it has been in the air for one-fourth of this time.</div>
<div class="partText">
<div class="partText">c.)Repeat for one-halfof the total time.</div>
<div class="partText">
<div class="partText">d.)Repeat for three-fourths of the total time.</div>
<div class="partText">
<div class="partText">e.)For each of these times, determine its speed. answer must be answered, numerically seperated by commas</div>
</div>
</div>
</div>
</div>
</p>
Explanation / Answer
Question a) By calculating sine of the angle, using right-triangle trigonometry, that the ball is kicked at, we can calculate the vertical velocity. This is the most important piece of information for calculating the total time. As the ground is flat (and I assume we can ignore air resistance), the time taken until the ball reaches its peak is the same as it takes to go from its peak back to the ground: sin(angle) = v_vert/v (v_vert is vertical velocity at take off, v is overall velocity at take off) sin(30) = v_vert/30 30sin(30) = v_vert v_vert = 15m/s At its peak, the ball is no longer moving in a vertical direction, so its velocity, vertically is zero. Using constant acceleration equations and assuming we are on Earth (a = 9.8ms^-2), we can calculate the time taken to reach the peak: v = u + at (v = final vertical velocity, u = initial vertical velocity, a = acceleration, t = time) 0 = 15 + (-9.8)*t (a is negative as it is acting in the opposite direction to the initial velocity) 9.8t = 15 t = 15/9.8 = 1.531s We can then double this to find the total time in the air: t_Total = 2*1.531 = 3.061s Question b) In order to find this angle, we need to know the vertical AND horizontal components of the velocity. As I am assuming there is no air resistance, there is no retarding force, or for that matter, any force at all on the ball, horizontally. Thus, the horizontal velocity is constant for the entire time the ball is in flight. We can use right-triangle trigonometry to find cosine of the initial flight angle (or Pythagoras' theorem) to find the horizontal velocity: cos(angle) = v_horiz/v (v_horiz = horizontal velocity, v = initial velocity) cos(30) = v_horiz/30 30cos(30) = v_horiz v_horiz = 25.981 m/s Now, to find the vertical velocity after one quarter of the total flight time, we can use constant acceleration equations: v = u + at (v = final vertical velocity, u = initial vertical velocity, a = acceleration, t = time) v = 15 + (-9.8)*3.061/4 v = 7.5 m/s Now that we have vertical velocity and horizontal velocity at one quarter of the flight time, by using right-triangle trigonometry, evaluating the arctangent, we will get the angle with respect to the horizontal: tan(angle) = v_vert/v_horiz tan(angle) = 7.5/25.981 tan(angle) = 0.2887 angle = arctan(0.2887) = 16.102 degrees Question c) This requires no calculation at all. At half of the total time, the ball is at its peak and instantaneously traveling at the tangent to its parabolic path. Thus, it is traveling parallel to the ground, or at an angle of 0 degrees relative to it. (If you want to do the calculation, the horizontal velocity is the same as calculated in part b, but the vertical velocity is 0). Question d) Again, this requires very little calculation. As the path of the ball is symmetrical about the peak, the angle the velocity makes at 3 quarters of the time is the supplement of the angle found in part b: Angle = 180 - 16.102 = 163.898 degrees Question e) We have already down all of the ground work for this. As we know the horizontal and vertical components of the velocity at each of these points, we only have to use Pythagoras' theorem to calculate the overall speed. i) At 1/4 of the flight, we have v_horiz = 25.981 and v_vert = 7.5 v^2 = v_vert^2 + v_horiz^2 v^2 = 7.5^2 + 25.981^2 v^2 = 56.25 + 675 = 731.25 v = sqrt(731.25) = 27.042 m/s ii) At 1/2 of the flight, we have v_horiz = 25.981 and v_vert = 0 Therefore, there is no calculation required for this one as the overall speed is just the horizontal speed: speed = 25.981 m/s iii) At 3/4 of the flight, we have v_horiz = 25.981 and v_vert = -7.5 As the only difference between this and 1/4 is the direction of the vertical velocity, the speed is exactly the same: speed = 27.042 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.