Question

A Ball Is Thrown Vertically Upword And It Reaches A Height Of 90M Find A) The Velocity Which It Was Thrown B) The Height Reached By The Ball 7 Second After It Was Thrown

Answer 1

Answer:

v = 0 [Ball is thrown upwards]

S = 90m [Given]

u = ? [We need to find u]

As ball is thrown upwards, gravity will act on it.

So, a = - 9.8 m/s²

A)

v² - u² = 2aS

0² - u² = 2 x -9.8 x 90

-u² = 180 x -9.8

-u² = -1764 [Negative sign gets cancelled]

u = √1764

u = 42 m/s

The velocity by which it was thrown is 42 m/s.

B)

u = 42 m/s

t = 7s

a = -9.8

S = Ut + 1/2at²

S = 42 x  7 + 1/2 x -9.8 x 49

S = 294 + -4.9 x 49

S = 294 - 24.01

S = 269.99

S ≈ 270 m

The height reached by the ball 7 seconds after it was thrown is 270 metres.

Answer 2

Given common data,

velocity of the object at maximum height $V= 0\frac{m}{s}$

Acceleration acting on the body is $g = (-9.81 \frac{m}{s^2})$

Note: Negative sign for acceleration is because it is acting downwards which in turn slows down the body.

Now,

A) velocity of ball when it was thrown is

Given,

maximum height reached by the object $s= 90m$

using kinematics equation

$V^2-U^2= 2*a*s$

by substituting the respective values we get

$0^2-U^2=2*(-9.81)*90\\\\-U^2=-1765.8\\ \\U^2 = 1765.8\\ \\U =42.02 \frac{m}{s}$

∴Velocity of the body when thrown is $U = 42.02 \frac{m}{s}$

B) Height reached by the ball in $7$ seconds after thrown.

Using the kinematic equation

given time $t =7sec$

$s=(U*t)+(\frac{1}{2} *a*t^2)$

by substituting the respective values we get

$S_t = (42.02*7)+(\frac{1}{2}*(-9.81)*(7^2))\\S_t = (294.14)-(240.345)\\S_t=53.795 m$

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