A projectile is fired from the ground level with a velocity of 300ms-1 at an angle of a 30o to the horizontal.
Calculate the time taken to reach the maximum height?
[g = 10ms-2]

Mightykay

8 years ago

time taken= 2Usin¤/2g

t= 2*300*0.5/20

t= 600*0.5/20

t= 300/20

t= 15secs

rasheedoyeladun

9 years ago

the answer is correct coz d question says time to reach d maxi height not total time of flight

Akabuezechristian

8 years ago

The answer is correct

Kingsley1145

4 years ago

I think the answer is not correct either is the question correct.
If we are to solve based on the question it means the answer is 1135m. Yes the unit of maximum height is in meter.. Then ur answer is wrong.

Then if we are to solve based on the answer u give out. That is to say we are solving for half the time it takes to reach the maximum height. Then the question is wrong..

Osiobe Fred

9 years ago

the answer is not correct

2usin@/g

Miss sparkle

9 years ago

The ans z 30s cos d time of flight formular is 2usin@/g.

Miss sparkle

9 years ago

Xori d ans z correct

Lillgen7562

9 years ago

2usin¤/g

U=300m\s

¤=30^•

g=10m\s^2

T= 2*300*sin30/10

T=600*0.5/10

T=300/10

T=30sec

Ceu Okaa

9 years ago

I was solving for max heights

=(U.sin0)*2/2g

e_musa

9 years ago

How is this ans correct

kingbest27

9 years ago

the correct answer is 15s which is d

Graceist

9 years ago

the quest says time to reach maximum height which is half of time of flight.time of flight is the total time taken to reach maximum height and come back...so d ans is correct

Toclay

5 years ago

See we guys we need to be careful,jamb is very dangerous,they said time taken to reach maximum height, not time of flight
We all know the time of flight but what of the time taken to reach maximum height
Just the half of Time of flight simple

Elbrain

9 years ago

usin(/)/g

kamal5

9 years ago

this answer is wrong. the formula of time of flight is 2usin@/10