The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t
This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²
This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting
Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Answer:
5 sq. root 3
Explanation:
theta= 60°
=> u sin theta = 10 × sin 60
= 10× sq. root 3/2
= 5 sq. root 3
Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
Answer:
ya we can write the imaginary character's name .
So that we can identify these imaginary people, as we cannot simply write the conversation and leave it .
Or maybe sometimes the reader will get confused as there is no name for the two people .
So, i suggest that you should write the names
Explanation:
You can even ask to your class teacher for further clarification
(a) Let's convert the final speed of the car in m/s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s: