Answer to Question #146709 in Physics for Althea

Question #146709
Suppose the body in Sample 20.0 is projected at 24.5 m/s, 30° above the horizontal. Find the a. time of flight, b. range, and c. Maximum height.
1
Expert's answer
2020-11-27T07:15:48-0500

Let's write the equations of motion of the body in horizontal and vertical directions:


x=v0tcosθ,(1)x=v_0tcos\theta, (1)y=v0tsinθ12gt2,(2)y=v_0tsin\theta-\dfrac{1}{2}gt^2, (2)


here, xx is the horizontal displacement of the body (or the range), v0=24.5 msv_0=24.5\ \dfrac{m}{s} is the initial velocity of the body, tt is the time of flight of the body, θ=30\theta=30^{\circ} is the launch angle, yy is the vertical displacement of the body (or the height) and g=9.8 ms2g=9.8\ \dfrac{m}{s^2} is the acceleration due to gravity.

a) Let's first find the time that the body takes to reach the maximum height from the kinematic equation:


vy=v0sinθgtrise,v_y=v_0sin\theta-gt_{rise},0=v0sinθgtrise,0=v_0sin\theta-gt_{rise},trise=v0sinθg.t_{rise}=\dfrac{v_0sin\theta}{g}.


Then, we can find the time of flight of the body:


t=2trise=2v0sinθg,t=2t_{rise}=\dfrac{2v_0sin\theta}{g},t=224.5 mssin309.8 ms2=2.5 s.t=\dfrac{2\cdot 24.5\ \dfrac{m}{s}\cdot sin30^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.5\ s.

b) We can substitute the time of flight of the body into the first equation and find the range of the body:


x=24.5 ms2.5 scos30=53 m.x=24.5\ \dfrac{m}{s}\cdot 2.5\ s\cdot cos30^{\circ}=53\ m.

c) Finally, we can substitute triset_{rise} into the second equation and find the maximum height:


ymax=v0sinθv0sinθg12g(v0sinθg)2,y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,ymax=v02sin2θ2g,y_{max}=\dfrac{v_0^2sin^2\theta}{2g},ymax=(24.5 ms)2sin23029.8 ms2=7.66 m.y_{max}=\dfrac{(24.5\ \dfrac{m}{s})^2\cdot sin^230^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=7.66\ m.

Answer:

a) t=2.5 s.t=2.5\ s.

b) x=53 m.x=53\ m.

c) ymax=7.66 m.y_{max}=7.66\ m.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog