Question

A spring-gun projects a small rock from the ground with speed v0 at an angle ?0 above the ground. You have been asked to determine v0. From the way the spring-gun is constructed, you know that to a good approximation v0 is independent of the launch angle. You go to a level, open field, select a launch angle, and measure the horizontal distance the rock travels. You use g = 9.80 m/s2 and ignore the small height of the end of the spring-guns barrel above the ground. Since your measurement includes some uncertainty in values measured for the launch angle and for the horizontal range, you repeat the measurement for several launch angles and obtain the results given in (Figure 1) You ignore air resistance because there is no wind and the rock is small and heavy.

a) Use the slope of the best straight-line fit to your data from the previous part to calculate v0.

b)When the launch angle is 35.9 ? , what maximum height above the ground does the rock reach?

Answer 1

Horizontal velocity = v_o cosθ_o

Vertical velocity = v_o sin θ_o

time of flight = 2 v_o sin θ_o/g

Range R =  v_o cosθ_{o *} 2 v_o sin θ_o/g = (v_o² sin 2θ_o)/g

So the slope of the R- Sin 2θ_o curve gives v_o²/g

=> v_o²/g = 11

=> v_o = 10.38 m/s

b) theta = 35.9 degree

vertical velocity intial, u =10.38 sin 35.9 = 6.0865 m/s

Maximum height = u²/2g = 1.89 m

Answer 2