Lab 3 | Physics homework help

Question 3: 1 point

Illustration and equations Decompose motion into two motions: Along y-axis: Motion with initial velocity ϑ 0 y = ϑ 0 sinθ under accelerationg . Along x-axis: Motion with initial velocity ϑ 0 x = ϑ 0cosθ with no acceleration. Studying the motions along the two axes Motion along y-axis: Follows Newton’s second law, as it has accelerationg owing to gravitational force acting downwards along y-axis. Governed by Eqn. (5): y = y 0 + (ϑ 0 sinθ)t− 2 g t 2 . Solving Eqn. (5), we find total time of flightT , which gives the stopping condition for overall projectile motion. NOTE: ThisT is different from the instantaneous time t . Motion along x-axis: Follows Newton’s first law, as there is no force in x-direction. Velocity in x-direction does not change. Range of flight R is given by using total flight timeT in Eqn. (4): R= (ϑ 0 cosθ)T . Experimental setup and procedure Measure initial shooting height and height of timing pad from ground to findy with ruler. Measure launching angle θ from the angular graduation on projectile launcher. Measure range of flight R with a tape measure. Measure initial velocity v 0 and the total flight time T with computer. Computation of ymax Motion in y-direction is guided only by gravity it is similar to→ throwing something upwards. Hence, the y-component of velocityϑ y becomes instantaneously 0 at topmost point of flight. Then, we have: ϑ y = 0 when y = ymax . Substituting this fact in Eqn. (3) : ϑ y 2 = ϑ 0 y 2 − 2g( y − y 0 ) we then obtain: 0 = ϑ 0 y 2 − 2 g( ymax − y 0 )

Solving for ymax , we then get:

ymax = y 0 + ϑ 0 y 2 g End of Theory Provided data for Exp 3 and instructions for data analysis and lab report