Do you have the answer choices ?
Answer:
(1) 
(2) Diagram has been attached in the solution.
Explanation:
This question is from projectile motion.
From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.
Part (1):
Let us assume:
= initial position of the basket ball = 2.1 m
= final position of the basket ball = 3.05 m
= acceleration of the ball along the vertical = 
= time taken to reach the goal = 0.8 s
= angle of the initial velocity with the horizontal
= initial speed of the ball = 7 m/s
= initial vertical velocity of the ball = u\sin \theta
Using the equation of motion for constant acceleration, we have
Hence, the angle of the shoot of the basket ball with the horizontal is such that it reaches the hoop on time.
Part (2):
For this part, a diagram has been attached.
#21
- initial velocity=u=5m/s
- Time=t=2s
- Acceleration=a=1.3m/s²
According to first equation of kinematics
- v=u+at
- v=5+1.3(2)
- v=5+2.6
- v=7.6m/s
#22
#a
- v1=2+3(3)=2+9=11m/s
- v2=-8-4(3)=-20m/s
- v3=1-5(3)=-14m/s
The order is
#b
for speed find absolute velocity
- S1=|11|=11m/s
- S2=|-20|=20m/s
- S3=|-14|=14m/s
So order is
S1<S3<S2
This depends entirely on the type of transformation of energy that is taking place, mechanical to thermal, thermal to mechanical, etc. But energy is always conserved, although it may take a different form.
