Question

A ball is thrown vertically upward with a speed of 19.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started?

Answer 1

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

Answer 2

Answer with Explanation:

We are given that

Initial speed=$u=19 m/s$

a.$g=9.8m/s^2$

Final velocity of ball=$v=0$

$v=u-gt$

g is negative because the ball is going against to gravity.

$0=19-9.8t$

$9.8t=19$

$t=\frac{19}{9.8}=1.94 s$

$s=ut-\frac{1}{2}gt^2$

Using the formula

$s=19(1.94)-\frac{1}{2}(9.8)(1.94)^2$

$s=18.4 m$

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

$s=ut+\frac{1}{2}gt^2$

$18.4=0(t)+\frac{1}{2}(9.8)t^2$

$18.4=4.9t^2$

$t^2=\frac{18.4}{4.9}$

$t=\sqrt{\frac{18.4}{4.9}}$

$t=1.94 s$

$v=u+gt$

Using the formula

$v=0+9.8(1.94)=19 m/s$