Question

A Ball is thrown vertically upwards with a velocity of 49m/s. Calculate

I) The maximum height to which it rises
II) The total time it takes to reach the surface of the earth​

Answer 1

Answer:

Explanation:

Vekh veere..

By conservation of energy

1/2mv^2=mgh

Putting values,

1/2 .49×49= 10h

h=2400/20= 120m

and we know

T=root(2h/g)=4.9s

Total time to reach earth surface,

9.8sec

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\bold{\therefore Height=122.5\:m}}$

$\green{\bold{\therefore Time\:taken=10\:sec}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{given :}} \\ \implies \text{initial \: velocity (u) = 49 \: m/s} \\ \\ \implies \text{acceleration(g) = - 9.8m}/s^{2} \\ \\ \implies \text{final \: velocity(v) = 0 m/s} \\ \\ \red{ \underline \bold{to \: find :}} \\ \implies \text{height(s) = ?} \\ \\ \implies \text{time \: taken (t) = ?}$

• According to given question :

$\bold{second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {0}^{2} = ({49})^{2} + 2 \times - 9.8 \times s \\ \\ \implies 0 - 2401 = - 9.8 \times 2 \times s \\ \\ \implies s = \frac{ - 2401}{ - 9.8 \times 2} \\ \\ \green{\implies \text{s = 122.5 m}} \\ \\ \bold{first \: equation \: of \: motion : } \\ \\ \text{from \: bottom \: to \: top}\\ \implies v = u + at \\ \\ \implies 0 = 49 + ( - 9.8) \times t_{1}\\ \\ \implies \frac{ - 49}{ - 9.8} = t \\ \\ \green{\implies {t_{1} = 5 \: sec}} \\ \\ \text{from \: top \: to \: bottom } \\ \implies s = ut + \frac{1}{2} {gt}^{2} \\ \\ \implies 122.5 = 0 \times t_{2} + \frac{1}{2} \times 9.8 \times {t_{2}}^{2} \\ \\ \implies \frac{122.5}{4.9} = {t}^{2} \\ \\ \implies t_{2}= \sqrt{25} \\ \\ \ \green{\implies {t_{2} = 5 sec}} \\ \\ \text{total \: time : } \\ \green{\implies t_{1} + t_{2} = 5 + 5 = 10 \: sec}$