An object is projected with a velocity of 50ms-1 from ground level at an angle θ to the vertical. If the total time of flight of the projectile is 5s, what is the value of θ? (g = 10ms-2)
0°
30°
45°
60°
90°
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Discussions (10)
very wrong they said to the vertical not horizontal it should be
T=2Usinθ/g
T=5s,u=50ms¹,g=10m/s²
5=2(50)cosθ/10
50=100cosθ
cosθ=1/2
θ=cos⁻¹0.5
=60°
D is correct
T=2UsinØ/g
T=5s,u=50ms¹,g=10
5=2(50)sin∅/10
50=100sin∅
sin∅=1/2
from tables,sin30⁰=1/2 or 0.5
=30⁰
One mistake students possible make us that they forget that they are asked to find the angle connecting the projectile path and the vertical.
So doing all calculations simplified
Total time of flight = 2usinθ/g
T=5 seconds, θ=? , g = 10ms-2, u = 50ms-1
5= 2*50sinθ/10
5= 100sinθ/10
5= 10sinθ
sinθ=5/10
sinθ= 0.5
θ = sin-1(0.5)
θ = 30
So the real θ inclined to the vertical is 90-30=60
the question stated that the angle was inclined to the vertical
hence to find the angle inclined to the vertical you have to first find the angle inclined to the horizontal using the formula T= 2usin0/g. then you subtract it from 90 degrees. the answer would be 60 degrees
REMEMBER WE WERE ASKED TO FIND THE ANGLE INCLINED TO THE VERTICAL!!!! WHICH IS 60
The angle inclined to the horizontal is the one that is 30 degrees
When the question is solved you get the angle to the horizontal as 30° but the question asked for the angle to the vertical which will then be 90°-30° = 60°
There by making the correct answer to be 60°
