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3 years ago

6

A child drops a ball from a window. The ball strikes the ground in 3 seconds. What is the velocity of the ball the instant befor

e it hits the ground?

1 answer:

nignag [31]3 years ago

6 0

Answer:

29.4 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3 s

Find: v

v = at + v₀

v = (9.8 m/s²) (3 s) + 0 m/s

v = 29.4 m/s

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A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on

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Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

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3 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago