Answer:
1D
2C
3C
4C
5D
6D
7D
8D
9B
Explanation:
better give me points X﹏X
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>

the velocity is Zero when the projectile reach in the maximum altitude:

When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>

R=Range


**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile

2B=60°
B=30°
Explanation :
Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.
Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.
When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.
When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)