A stone is projected at an angle of 60° to the horizontal with velocity of...

A stone is projected at an angle of 60° to the horizontal with velocity of 30m/s. Calculate,(a) the maximum height (b) the range (c) Time taken to reach the maximum height (d) Total time of flight (e) the height of the stone at the instant that the part make an angle of 30° horizontal.?

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Answers (3)

Jay-Jay
3 months ago
@=60
v=30m/s
g=10m/s2
a)maximum height
H=v^2cos^2@/2g
H=30^2 (cos60)^2/2*10
H=900*(0.5)^2/20
H=45*0.25
H=11.25m

b)range
R=v^2 sin2@/g
R=30^2 sin2(60)/10
R=900*0.8266/10
R=90*0.866
R=77.94

c)time taken
t=vcos@/g
t=30cos60/10
t=1.5sec

d)T=2t
T=2*1.5
T=3sec

e)at 30
H=v^2 cos^2 @/2g
H=30^2 cos^2 (30)/2*10
H=900*(0.866)^2/20
H=45*0.75
H=33.75
scientist
3 months ago
33.75m
scientist
3 months ago
v=30m/s
g=10m/s2
a)maximum height
H=v^2cos^2@/2g
H=30^2 (cos60)^2/2*10
H=900*(0.5)^2/20
H=45*0.25
H=11.25m

b)range
R=v^2 sin2@/g
R=30^2 sin2(60)/10
R=900*0.8266/10
R=90*0.866
R=77.94

c)time taken
t=vcos@/g
t=30cos60/10
t=1.5sec

d)T=2t
T=2*1.5
T=3sec

e)at 30
H=v^2 cos^2 @/2g
H=30^2 cos^2 (30)/2*10
H=900*(0.866)^2/20
H=45*0.75
H=33.75
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