Answer:
117,30m
Explanation:
I think the situation here is a horizontal projection to the ground. So in order to find the distance the formula = Ut, where U is the initial speed and t is the Time of flight. To get the time of flight in this case =√2h/g where h is the height and g is gravity. so to get the time = √2×300÷9.81 =7.821 .so range =ut which is equal to the time multiplied by 15m/s =117.30m
Answer:
Explanation:
we can look for the final velocity of the object using the eqaution of motion as shown:
v² = u²+2gH
v is the final velocity
u is the initial velocity = 10m/s
g is the acceleration due to gravity = 9.81m/s²
H is the height of the object = 175m
Subxtitute the given parameters inti the formula and get v:
v² = 10²+2(9.81)(175)
v² = 100+3433.5
v² = 3533.5
v = √3533.5
v = 59.44m/s
Hence the final velocity of the object is 59.44m/s
The position of the sun must be at the horizon in order for the rainbow to appear.
The relative positions of the sun and plane make the rainbow appear as a full circle.
However, if the sun is at the horizon, the shadow of the plane will be behind the plane, not below it; this means that the shadow of the plane will not fall on the earth.
Answer: h = 20.92 m
Explanation: By using the law of conservation of energy, the kinetic energy of the ball equals it potential energy.
Kinetic energy =mv^2/2
Potential energy = mgh
Where m = mass of the object, v = velocity of object = 23.5 m/s
g = acceleration due gravity = 9.8 m/s^2
mv^2/2 = mgh
m cancels out each other on both sides , hence we have that
v^2 = 2gh.
We want the ball to move towards the wall (horizontal motion), hence we need the horizontal component of the velocity since the velocity is inclined at an angle of 30.5 to the ground (horizontal).
Hence v = 23.5 × cos 30.5, v = 20.248 m/s
Recall that v^2 = 2gh
(20.248)^2 = 2×9.8×h
409.98 = 19.6 h
h = 409.98/ 19.6
h = 20.92 m
