Question

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point

Answer 1

Heya!

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(i)

v= 0 (at highest point, the velocity is zero)
u=?
s=10m
a= -10 m/s^2 ( object is thrown against the gravity)

Using the second equation of motion,

v^2-u^2=2as
0-u^2= 2×-10×10
0-u^2= -200
-u^2= -200
u= √200
u= 10√2 m/s
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(ii)

t=?

Using the first equation of motion,

v-u= at
0-10√2= -10×t
-10√2= -10×t
10√2=10t
10√2/10=t
t=√2 s

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Hope it helps...!!!

Answer 2

Given- Distance travelled (s) = 10 m
Final velocity (v) = 0m/s
g = - 9.8 m/s^2

To find- (i) Initial velocity
(ii) Time taken

Solution :
(i) By Third Equation of Motion,
$v {}^{2} = u {}^{2} + 2 \: as \\ \: 0 = u {}^{2} + 2 \times ( - 9.8 \: ms {}^{ - 2} ) \times 10 \: m - u {}^{2} \\ \: \: \: \: \: = - 2 \times 9.8 \times 10 \: m {}^{2} s {}^{ - 2} \\ \: u = \sqrt{196 } \: ms {}^{ - 1} \\ \: u = 14 \: ms {}^{ - 1}$

(ii) By First Equation of Motion,
$v = u + at \\ 0 = 14 \: ms {}^{ - 1} - 9.8 \: ms {}^{ - 2} \times t \\ t = 1.43 \: s$
Thus,
(i) Initial velocity, u = 14 m/s
(ii) Time taken, t = 1.43 s.