Answer to Question #139501 in Physics for edison

Question

A physics student launches a projectile from the surface of a table at an angle of 30.0 degrees. It lands 2.00 m away on the same table. (a) What is the initial speed of the projectile? (b) What is the time of flight? (c) What would the time of flight and initial velocity be if the launch angle was 60.0 degrees instead?

Answer 1

a)

$$R=\frac{v^2\sin{(2\cdot30)}}{g}\\2=\frac{v^2\sin{60}}{9.8}\\v=4.76\frac{m}{s}$$

b)

$$t=\frac{2v\sin{30}}{g}=\frac{2(4.76)\sin{30}}{9.8}=0.486\ s$$

c)

$$R=\frac{v'^2\sin{(2\cdot60)}}{g}\\2=\frac{v'^2\sin{120}}{9.8}\\v'=v=4.76\frac{m}{s}$$

$$t'=\frac{2v\sin{60}}{g}=\frac{2(4.76)\sin{60}}{9.8}=0.841\ s$$