Use the formula for distance dropped from rest:
D = 1/2 a t^2 .
With no air resistance, you get about 8.8 sec.
I'm pretty sure you can fill in the missing work.
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d =
×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d =
=
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
Answer:
The maximum range is 300% of the range of the projectile is projected at an angle of 9.74°.
None of the options are correct.
Explanation:
Normally, ignoring air resistance, for projectile motion, the range (horizontal distance travelled) of the motion is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched
g = acceleration due to gravity = 9.8 m/s²
The range is maximum when θ = 45°
R when θ = 45° is
R = (u² sin 90°)/g = (u²/g)
We are then told that this maximum range is 300% of the value obtainable for the range at a particular angle
Maximum range = 3R
(u²/g) = 3(u² sin 2θ)/g
Sin 2θ = (1/3)
2θ = sin⁻¹ (1/3) = 19.47°
θ = (19.47°/2) = 9.74°
Hope this Helps!!!
Answer:
c
Explanation:
When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.
You can view in the attached diagram to understand how the velocity is being changed.