A ball is thrown with speed 20m/s at an angle 30° with vertical. Its height above the point of projection when it is at horizontal distance of 5m from the thrower is (g=10m/s)
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Range is 10m
horizontal velocity
=ucos30
=20× (√3 /2)
=10√3 m/s
vertical velocity
= u sin30
= 20x 1/2
= 10 m/s
Time
= 2u/g
= 20 x 2 /10
= 4
hight
=( u² sin² 30 )/2g
= 400 /4×2 ×10
= 5m
Range
=u² sin2x30
=400 √3 /2
= 200√3
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horizontal velocity
=ucos30
=20× (√3 /2)
=10√3 m/s
vertical velocity
= u sin30
= 20x 1/2
= 10 m/s
Time
= 2u/g
= 20 x 2 /10
= 4
hight
=( u² sin² 30 )/2g
= 400 /4×2 ×10
= 5m
Range
=u² sin2x30
=400 √3 /2
= 200√3
I hope u understand
Am I right?
then please follow me
and mark me brainlist
thank you!!!
annika0:
A ball is thrown with speed 20m/s at an angle 30° with the vertical. Its height above the point of projection when it is at horizontal distance of 5m from the thrower is ( g = 10m/s
a) 7.41m b) 8.41m c)6.14m d)5.14m
Try to solve & please help me out
ok
wait
Yeah sure Thank u
t = 2 / √3
no
t= 1/ 2√ 3 ,........... then.........H =2.88m
hi
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