Question

(1 point) A ball is thrown at 35A below the horizontal from a rooftop of height 65 m. It lands 2.7 s later. Find: Find a) its initial speed; 18.89 m/s b) the point of impact with the ground; 41.78 m from the base of the building c) the angle at which it hits the ground. Å below the horizontal

Answer 1

NT K) 35 The ball takes 2.7 Sre to land. 6m Mow, n= ut + 1 gth 65m on, 65 - W2. (2 В 65 = 2.7 +1x .8 2 0 41.78 m B on, UE 65 - 35.721 on, u = 217 - 10.844 ms. component of initial speed which is the vertical so, uxsin 35° = 10.844 1. 10.844 (a) on, ua 0.574 - 18.89 mis. (6) Now, Honizontal range = Unx to Henizontal velocity = Un = u. cos 35° 18.89 x 0.819 15. 471 ms. so range = 15.471 X 2.7 = 41.78 metne.

(0) Horizontal velocity remains constant Throughout the NOW motion. Remains constant At B, V. = 15.491 mis VyUvt at 10.849 + 9.8 x 2.7 ms = 37. 304 ms.

So, tan ß = - 37. 304 15.471 MENNYT se, ton pa se on, B = tani(2.11). or, angle B = 67. 47º. is 67 47° below the honizontal I so, the