Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.
Answer:
96 m
Explanation:
Given,
Initial velocity ( u ) = 4 m/s
Final velocity ( v ) = 20 m/s
Time ( t ) = 8 s
Let Acceleration be " a ".
Formula : -
a = ( v - u ) / t
a = ( 20 - 4 ) / 8
= 16 / 8
a = 2 m/s²
Let displacement be " s ".
Formula : -
s = ut + at² / 2
s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2
= 32 + ( 2 ) ( 64 ) / 2
= 32 + ( 2 ) ( 32 )
= 32 + 64
s = 96 m
Therefore, it travels 96 m in time 8 s.
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
