Answer with Explanation:
We are given that
Initial speed=[tex]u=19 m/s[/tex]
a.[tex]g=9.8m/s^2[/tex]
Final velocity of ball=[tex]v=0[/tex]
[tex]v=u-gt[/tex]
g is negative because the ball is going against to gravity.
[tex]0=19-9.8t[/tex]
[tex]9.8t=19[/tex]
[tex]t=\frac{19}{9.8}=1.94 s[/tex]
[tex]s=ut-\frac{1}{2}gt^2[/tex]
Using the formula
[tex]s=19(1.94)-\frac{1}{2}(9.8)(1.94)^2[/tex]
[tex]s=18.4 m[/tex]
a.The ball rise upto height 18.4 m
b.It take 1.94 s to reach its highest point.
c.Initial velocity=0,s=18.4 m
[tex]s=ut+\frac{1}{2}gt^2[/tex]
[tex]18.4=0(t)+\frac{1}{2}(9.8)t^2[/tex]
[tex]18.4=4.9t^2[/tex]
[tex]t^2=\frac{18.4}{4.9}[/tex]
[tex]t=\sqrt{\frac{18.4}{4.9}}[/tex]
[tex]t=1.94 s[/tex]
[tex]v=u+gt[/tex]
Using the formula
[tex]v=0+9.8(1.94)=19 m/s[/tex]
Answer:
Explanation:
initial speed, u = 19 m/s
(a) Let it rises upto height h.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity and it is zero.
0 = 19 x 19 - 2 x 9.8 x h
h = 18.4 m
(c) Let the ball takes time t to reach to the maximum height.
use first equation of motion
v = u - gt
0 = 19 - 9.8 x t
t = 1.94 s
(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height
T = 2 x t = 2 x 1.94 = 3.88 s
(d) When the ball reaches the ground, let the velocity is v.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity
v² = 0 + 2 x 9.8 x 18.4
v = 19 m/s
