Answer:
height from where rock was thrown is 27.916 m
Explanation:
speed = 7.50 m/s
θ = 30°
g= 9.8 m/s²
horizontal distance = 18 m
time require for vertical displacement

t = 2.8 sec
now for calculation of height
s = ut + 0.5 a t²
-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)
-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)
-h = -27.916 m
h= 27.916 m
height from where rock was thrown is 27.916 m
Answer:
D). energy resulting from the attraction between two masses.
Explanation:
D. Number of cycles/ unit of time
Explanation:
It is given that,
Height above which the stone was thrown, h = 10 m
Initial velocity of the stone, u = 8 m/s
Angle above the horizontal, 
The horizontal component of velocity is, 
The vertical component of velocity is, 
Let t is the time of flight in vertical motion. The second equation of motion is :

t = 0.34 seconds
Let s is the range of the stone. It can be calculated as :


s = 2.46 meters
So, the range of the stone is 2.46 meters. Hence, this is the required solution.