Question

A stone is thrown vertically upward with a speed of 28 m/s.

(a) Find the maximum height reached by the stone.

(b) Find its velocity one second before it reaches the maximum height.

(c) Does the answer of part

(d) Change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s ?

Answer 1

Given, u = 28 m/s, v= 0, a = -g = -9.8 $m/s^2$

$S_2=\frac{v_2-u_2}{2a}$

=$\frac{0_2-{28}_2}{2(-9.8)}=40m$

(b) time, t =\( \frac{(v-u)}{a}) = \frac{0-28}{-9.8} = 2.85\
)

t' = 2.85 - 1 =1.85

v' = u +at' = 28 - (9.8) (1.85)

= 28-18.13 =9.87 m/s

Hence the velocity is 9.87 m/s.

(c) No it will not change. as, after one second velocity becomes zero for any initial velocity and deceleration is g = 9.8 \
(m/s^2\) remains same. For initial velocity more than 28 m/s maximum height increases.