13. Tr = 6/2 = 3 s. = Rise time.
Tf = Tr = 3 s. = Fall time.
a. V = Vo + g*Tr = 0 @ max h.
Vo = -g*Tr = -(-9.8)*3 = 29.4 m/s. =
Initial velocity.
b. h = Vo*Tr + 0.5g*Tr^2.
h = 29.4*3 - 4.9*3^2 = 44.1 m.
c. Since Tr is 3 seconds, the ball is
on its' way down after j seconds: Tf = 4-3 = 1 s.
h = ho-0.5g*Tf^2 = 44.1 - 4.9*1^2 = 39.2 m. Above launching point.
13. A ball is thrown vertically upwards and it returns to the thrower after 6 sec ( g=9.8 m/s2) Find.
(i) The velocity with which it was thrown up. (ii) The maximum height it reaches
(iii) Its position after 4 sec.
1 answer
1 answer