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Lerok [7]
3 years ago
15

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

What was the pop-up velocity of the ball?
Physics
1 answer:
kondaur [170]3 years ago
6 0

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s

Initial pop up velocity is 35.87 m/s

a)

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s

It took 3.65 seconds to reach this height

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A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct
-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

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4 0
3 years ago
You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

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the angle of incidence θ is 45.56 º

4 0
3 years ago
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