A ball is thrown vertically upwards with a velocity of 20 metre per second inverse from the top of the multi storey building the height of the point from where the ball is thrown is 25.0 metre from the ground (A)how h i g h will the ball rise ? and (b) how long will it be before the ball hits the ground? Take g equal to 10 metre per second square?

Question

Answer

Height of building=25m,
u=20m/s(+ve as in +ve y-axis),
v=0m/s,
a=g=-10m/s²,
a)h=0²-(-20)²/2(-10)=-400/-20=20m,
b)t taken to cover total distance of d=20m,
V=Vo+at,
0=20-10t,
t=2s,
for
T.Distance=45m
y=0m/s,
yo=45m,
g=-10m/s²
t=3s,
T.time=2+3=5s

Answer

Problem statement

A ball is thrown vertically upwards with a velocity of 20 metre per second inverse from the top of the multi storey building. The height of the point from where the ball is thrown is 25.0 metre from the ground.

Solution

Part (a)

To find the maximum height reached by the ball, we can use the kinematic equation:

$v^2 = u^2 + 2as$

where,

v = final velocity = 0 m/s (at maximum height, the ball will momentarily come to rest)
u = initial velocity = 20 m/s
a = acceleration due to gravity = -10 m/s^2 (since the ball is moving upwards, the acceleration is in the opposite direction to the velocity)
s = displacement = ?

Substituting the values in the equation, we get:

$0^2 = 20^2 + 2(-10)s$

Solving for s, we get:

s = 20 m

Therefore, the ball will rise to a height of 20 metres.

Part (b)

To find the time taken by the ball to hit the ground, we can use the kinematic equation:

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

where,

s = displacement = 25 m (since the ball is thrown from a height of 25 m)
u = initial velocity = 20 m/s
a = acceleration due to gravity = 10 m/s^2 (since the ball is moving downwards, the acceleration is in the same direction as the velocity)
t = time taken = ?

Substituting the values in the equation, we get:

$25 = 20t + \frac{1}{2}(10)t^2$

Simplifying the equation, we get:

$t^2 + 4t - 5 = 0$

Solving for t using the quadratic formula, we get:

$t = 1$ s or $t = -5$ s (we can ignore the negative value since time cannot be negative)

Therefore, the ball will hit the ground after 1 second.

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