Question 3 4m T T - 130 m/s ht4 At maximum height 'h' final speed (v)=0 initial speed (u) = + 30 m/s. g=10m/s² V²ut 2gH 0 = (+30) ²2x10xh - 900=-20th Downward direction as-ve. upward direction as tve. Maximum height from the ground y = 45m²4m y = 49m h=+45m (maximum Height) from initial position

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter9: Relativity
Section: Chapter Questions
Problem 2CQ: Explain why, when defining the length of a rod, it is necessary to specify that the positions of the...
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What is the velocity of the bullet the instant the bullet hits the ground ?

thiz
folloi Question 3
A:
2 = 1 4m
10)
L
öld,
T
130 m/s
0
1
At maximum height 'h'
final speed (V)=0
initial speed (M) = + 30 m/s.
g==10m/s²
и
√ ² u²t 2gH
0 = (+·30) = 2x10xh
-900=-20th
Question 4 h-4m
g=10m/s²
h +4
By using the equation
of motion
h=+45m (maximum Height)
from initial position
Downward direction as-ve.
3 upward direction as tve.
2
Maximum height from the ground
y= 45 mt4m.
y = 49m
4= 30m/s
2
S=ut that²
2
~ 4m=30++ √₂x10 +²
- 5+² +30 ++4-0
+ = 6.13s 3 + = -0.135
Transcribed Image Text:thiz folloi Question 3 A: 2 = 1 4m 10) L öld, T 130 m/s 0 1 At maximum height 'h' final speed (V)=0 initial speed (M) = + 30 m/s. g==10m/s² и √ ² u²t 2gH 0 = (+·30) = 2x10xh -900=-20th Question 4 h-4m g=10m/s² h +4 By using the equation of motion h=+45m (maximum Height) from initial position Downward direction as-ve. 3 upward direction as tve. 2 Maximum height from the ground y= 45 mt4m. y = 49m 4= 30m/s 2 S=ut that² 2 ~ 4m=30++ √₂x10 +² - 5+² +30 ++4-0 + = 6.13s 3 + = -0.135
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