Question

2.
A particle is projected vertically upward with a speed of 70 m/s from ground. Height of particle at t = 3 s and
t = 11 s are h, and he respectively. The value of (h. + ha) is (g = 10 m/s)
(1) 165 m
(2) 720 m
Me
(3) 680 m
(4) 330 m​

Answer 1

Answer:

s =70m,u=50,a=−10

using:s=ut−

2

gt

2

70=50t−5t

2

5t

2

−50t+70=0

t

1

& t

2

are two roots of above equation

then smaller root is :time when it crosses 70m

larger root is : time when it comes back below 70m

t

2

−t

1

=

a

b

2

−4ac

=

5

((−50)

2

−4×5×70

=2

11

=6.63sec

Answer 2

Explanation:

initial velocity , u = 70 m/s

acceleration due to gravity , g = - 10 m/s^2

time , t = 3 s

Using 2nd Equation of motion ,

$h = ut + \frac{1}{2}g {t}^{2}$

$= 70(3) - \frac{1}{2} \times 10 \times 9$

$= 210 - 45 = 165 \: m$

When t = 11 s

$ha = 70(11) - \frac{1}{2} \times 10 \times 121$

$= 770 - 605 = 165 \: m$

Then the value of h + ha = 165 + 165 = 330 m

I hope it helps you. If you have any doubts, then don't hesitate to ask .