pls answer this question
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v=30m/s
u=0 m/s
a=10m/s²
t=1.5 s
Then sin A = 1.5×10/30 = 15/30= 1/2
A= sin inverse(0.5)
A= 30°
Range =v² sin 2A/a
Range =(30)² sin 2(30)° / 10
= 900sin 60°/10
= 900 √3/2/10
= 450√3/10
= 45√3
u=0 m/s
a=10m/s²
t=1.5 s
Then sin A = 1.5×10/30 = 15/30= 1/2
A= sin inverse(0.5)
A= 30°
Range =v² sin 2A/a
Range =(30)² sin 2(30)° / 10
= 900sin 60°/10
= 900 √3/2/10
= 450√3/10
= 45√3
MOSFET01:
also in my ans m
tq bro
welcome
no bro
thats not m^2 i tried erase that
thats m only
Answered by
1
it is very simple i will give u a relation between RANGE AND MAXIMUM HEIGHT
U can solve many problems with that
relation between R and H
R/H =4cot(theta)
{theta = angle between velocity and horizontal
U can solve many problems with that
relation between R and H
R/H =4cot(theta)
{theta = angle between velocity and horizontal
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45 √3 m
:-)
m² replace by m
tq bro
bro can u solve the problem
??
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