Question #168476

GO TO THIS LINK:

https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html


SOLVE RANGE, TIME, MAX HEIGHT AND FINAL SPEED/VELOCITY BY THE GIVEN BELOW.

CANON HEIGHT CANON ANGLE INITIAL VELOCITY

 0 meters 0⁰ 15 m/s

0 meters 45⁰ 15 m/s

0 meters 70⁰ 15 m/s

15 METERS -5⁰ 15 m/s

15 METERS 0⁰ 15 m/s

15 METERS 45⁰ 15 m/s

15 METERS 70⁰ 15 m/s


THANKS. HELP ME WITH THIS ONE WITH SOLUTION



1
Expert's answer
2021-03-04T15:19:08-0500

a)


R=152sin09.8=0 mt=215cos09.8=3.1 sh=152cos2029.8=11.5 mv=15msR=\frac{15^2\sin{0}}{9.8}=0\ m\\ t=2\frac{15\cos{0}}{9.8}=3.1\ s\\ h=\frac{15^2\cos^2{0}}{2\cdot9.8}=11.5\ m\\v=15\frac{m}{s}

b)


R=152sin909.8=23 mt=215cos459.8=2.2 sh=152cos24529.8=5.7 mv=15msR=\frac{15^2\sin{90}}{9.8}=23\ m\\ t=2\frac{15\cos{45}}{9.8}=2.2\ s\\ h=\frac{15^2\cos^2{45}}{2\cdot9.8}=5.7\ m\\v=15\frac{m}{s}


c)


R=152sin1409.8=14.8 mt=215cos709.8=1.0 sh=152cos27029.8=1.3 mv=15msR=\frac{15^2\sin{140}}{9.8}=14.8\ m\\ t=2\frac{15\cos{70}}{9.8}=1.0\ s\\ h=\frac{15^2\cos^2{70}}{2\cdot9.8}=1.3\ m\\v=15\frac{m}{s}

d)


R=24.2 mt=1.6 sh=15 mv=22.8msR=24.2\ m\\ t=1.6\ s\\ h=15\ m\\v=22.8\frac{m}{s}

e)


R=26.2 mt=1.7 sh=15 mv=22.8msR=26.2\ m\\ t=1.7\ s\\ h=15\ m\\v=22.8\frac{m}{s}

f)


R=33.3 mt=3.1 sh=15 mv=22.8msR=33.3\ m\\ t=3.1\ s\\ h=15\ m\\v=22.8\frac{m}{s}

g)


R=19 mt=3.7 sh=25.1 mv=22.8msR=19\ m\\ t=3.7\ s\\ h=25.1\ m\\v=22.8\frac{m}{s}


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