Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
Answer:
the force exerted by the seat on the pilot is 10766.7 N
Explanation:
The computation of the force exerted by the seat on the pilot is as follows:

Hence, the force exerted by the seat on the pilot is 10766.7 N
Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
Answer:
Option (b) is correct.
Explanation:
The motion under the influence of gravity is called projectile motion.
The acceleration due to gravity is constant through out the motion and it is always acting downwards.
When an athlete jumps and follow the projectile path, it always have the same horizontal velocity as there is no acceleration in the horizontal direction.
Also he has the vertical acceleration constant which is equal to the acceleration due to gravity and acts towards the center of earth.
Option (b) is correct.