Question

a ball is throw vertically upward with a velocity of 30ms¹ from the top of a building the height of the point from where the ball is thrown 30m from the ground 1) how high will the ball rise
2) when ball reaches the ground from a height​

Answer 1

Explanation:

here

${v }^{2} = {u}^{2} - 2as$

=u²- 2 g h

or 0= 30² -2×10×h

so, h= 900/20

=45 m

so the ball will rise 45 m from top of building.

now total distance from ground is 45 +30 m= 75 m

g = 10m/s2

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

75= 0× t +1/2 . 10 t²

t²= 75/5= 15

or t = √15 s

Answer 2

Topic :- Motion in Staright line

Subtopic :- Motion under gravity

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

A ball is thrown vertically upwards from the top of a building with a velocity of 30 m/s i.e initial velocity (u) will be 30 m/s. The height of the point from where the ball is thrown 30 m from the ground and Acceleration due to gravity (g) will be -9.8 m/s² because its direction is towards downwards.

$\maltese\:\underline{\textsf{ Maximum height covered by the Particle :}}\:\maltese$

$:\implies \sf H = \dfrac{u^2}{2g}$

$:\implies \sf H = \dfrac{ {(30)}^{2} }{2 \times 9.8}$

$:\implies \sf H = \dfrac{ 900}{2 \times 9.8}$

$:\implies \sf H = \dfrac{900}{19.6}$

$:\implies \underline{ \boxed{\sf H =45.9 \: m }}$

$\maltese\:\underline{\textsf{Total time of flight :}}\:\maltese$

$\longrightarrow\:\:\sf T = \dfrac{2u}{g}$

$\longrightarrow\:\:\sf T = \dfrac{2 \times 30}{9.8}$

$\longrightarrow\:\:\sf T = \dfrac{60}{9.8}$

$\longrightarrow\:\: \underline{ \boxed{\sf T = 6.12 \: s }}$