Question

THE EXPERIMENT The speed "." of a ball as it leaves the spring gun will be measured by projecting the ball horizontally (0-0) from a table top at a distance H above the floor. The gun is then elevated at a known angle 8. The position of the target is calculated using the measured value of v. calculated time of flight, and the equations of motion. A successful prediction with the projectile hitting the target means that the following concepts about projectile motion are valid: The horizontal component of the motion is independent of the vertical component in the two dimensional motion; 2. Since there is no net horizontal force acting on the projectile, there is no horizontal component of the acceleration. This provides an experimental proof of Newton's 2 law. 3. If a 0, once motion is imparted to the projectile, it will continue to move such that the horizontal component of its velocity is constant in time. This is a consequence of Newton's first law: "An object at rest remains at rest; while an object in motion will continue to move with the same velocity unless acted upon by a net (unbalanced) force." L. HORIZONTAL SHOT The initial projecting velocity v, can be measured by firing the spring gun horizontally: From Eq. 4, X-R is the horizontal distance the projectile travels from the launch point to the floor. V. cos 0 becomes vo(1) because 0-0. The time is not measured but can be obtained by comparing the vertical displacement and time of flight of the projectile that is launched horizontally and an object that is dropped from rest. It is required that both start from the same height and have the same vertical displacement H. In effect, the time for the projectile to hit the floor can be obtained from the vertical part of the motion. Denote the projectile by the subscript'p' and the object that is dropped from rest by the prime symbol. Choose the origin (0,0) at the launch point which is 1 meters above the floor. [With this choice, what is the y coordinate of the floor?) For the projectile which is fired horizontally, Vpoy and the y motion is given by Eq. 6(a) Yp - 0 - (1/2)gt, Eq.8(a) For the object that is dropped from rest, Voy = 0 and its y displacement is y'-0-(1/2)gt" Eq.8 (b) Since both travel the same distance, yoy then, t' and the time of flight for the horizontal projectile is calculated from Eq.8 (a) or Eq. 8(b).

Answer 1

Given that 100) L (-242,-3) Dan 2 Bocedure X (honsonstel displacment or Rayer) when the ball strikes I the floor is duam. ... Y coordinate of the bill when it hits Fluore Bim . -. in Co-ordinate system it can be witten as (-243, -81m) No matter one can through the ball in fight side . allo in that case Cardinate will be ( 243, -81) ② y Gordinate will be the resuely motion enda ravity of veilical component for veliced Componet U=0 S- utt Lat2 (+81)= O(+ 1x 9.8*42 +2. 8172 to 4.06 sec . 9-8

. - for eflladian in x-direction Some through all motion to horizontal relooty xmain L isa ütt liat211 . Horizontal displacimal in | 243 = 64406) + I lojt2 lbecause horizontal - Kelocity const so Uz 242 : 04.06 59.76 mis xeraction to find voisi ha Volt 6 voz sanomis) as discreved earlier SALO a llo SiOS- olesist Andle shots Ha bsm R: 291m. R = 18 - 291- 0 x eflation for congle Shot TR = (hoceso) t 7. o Where Ro Range : Ozinclined anle Bt total time of flight

leplertion for an angle of shot it Wşa utt data in H= (lo sino)'t+ } l-g}{2 - Where He Hight=&sm. to total time of Flight Yvalue will be just the displacmed inveticalle.&sm ③ b-ie sign shows below the origin 9 Time of flight for an ayle shot i Stop R putting in en ② n locoso. He Closing) R1(s) (Red Rtano - 1 g R2 (locale
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65=291 ton is' – 0:5X98 x(29192 laz (Cos 18')2 85 = 291X0.2249 – 0.5X9-8 x(29132 ²x0.900 0.5x9.84(2912 291X0:32U9 - 85 = Vo? (0.904 j 9.5459 x 0.904 is 0.5*9.8 X(291)2 8.62 = uiua 36.9 4149369 8,62 Uo= 219.4 mis So to R 2291 Klocoso (214.4) (Cos1 8°) It= 1.39 sec.