Question

3. A ball is thrown from ground level. 3.00 s later it is moving horizontally at 15.0 m/s. Find: (a) the horizontal range; (b) the angle of impact.

Answer 1

ANSCOBR) Giver: SA Velocity = 15 mis avy time to see a Ball was moving implies that horizontally at to see has reache maximum height.

So, time for total Journey, T= 2t 2x3 65. a) Range, R-VT. (15) x (6 40 mm va b) Direction is given by orton ( 2 ) so from equation - Vyt at & Vy= -at find vertel ttelowly das ag =-418 mis? 80, Veyi-(-9.8) (3) = 29.4 m/s

Answer 2

Jg TAR=35 Vocusa=15misl B Vocoso A R @ Horizontal range, R = V2² sin zo where, tan= rosina = 3 -0 g ③ and Vocosa = 15- from eg. O, ② & ③ & Rg=avo sino Volosa Rg= 2(39) 15 [R = gom] ③ angle of impact from eg. & ② and = 3 7 tana = 3x9.8) 12=6307