We can solve this problem using the equations of motion under constant acceleration due to gravity (assumed to be -9.8 m/s^2):
i. The time of flight is given by:
t = 2 * (v0 * sinθ) / g
where v0 = initial velocity = 30 m/s, θ = angle with horizontal = 60°, and g = acceleration due to gravity = -9.8 m/s^2.
Substituting the values, we get:
t = 2 * (30 * sin60°) / 9.8
t = 3.06 seconds
ii. The horizontal displacement is given by:
d = v0 * cosθ * t
where cosθ = cos60° = 0.5.
Substituting the values, we get:
d = 30 * 0.5 * 3.06
d = 46.44 meters
iii. The maximum height reached by the ball can be found using:
h = (v0^2 * sin^2θ) / (2 * g)
Substituting the values, we get:
h = (30^2 * sin^2(60°)) / (2 * 9.8)
h = 45.92 meters
iv. The time taken to reach the maximum height can be found using:
t_max = v0 * sinθ / g
Substituting the values, we get:
t_max = 30 * sin60° / 9.8
t_max = 1.53 seconds
The tennis ball was released from a shooting pot with the initial velocity of 30m/s at
the angle of 60° with the horizontal.
Determine:
i. The time of the flight
ii. The horizontal displacement
iii. The maximum height the ball reached
iv. The time to reach maximum height
5 answers
2) A car is moving in curved bank ,the 900kg car is moving at speed of 40.0 m/s through
this curve of radius 100m, a) what does the angle θ has to be if the road is frictionless?
b) If a car moving at constant speed, calculate friction force the is experiencing
this curve of radius 100m, a) what does the angle θ has to be if the road is frictionless?
b) If a car moving at constant speed, calculate friction force the is experiencing
a) If the road is frictionless, the centripetal force required to keep the car moving in the circular path is provided entirely by the normal force (since there is no friction). The centripetal force is given by:
F_c = m * v^2 / r
where m = mass of the car = 900 kg, v = speed of the car = 40.0 m/s, and r = radius of the curve = 100 m.
Substituting the values, we get:
F_c = 900 * 40.0^2 / 100
F_c = 144000 N
Since the normal force N is perpendicular to the horizontal surface, we can use trigonometry to find the angle θ.
The vertical force balance equation is:
N - mg = 0
where g = acceleration due to gravity = 9.8 m/s^2. Rearranging, we get:
N = mg = 900 * 9.8 = 8820 N
The horizontal force balance equation is:
F_c = N * sinθ
Substituting the values, we get:
144000 = 8820 * sinθ
sinθ = 16.33
However, this is not a valid answer since we cannot have a sine greater than 1. This means that it is not possible to have a frictionless road that allows the car to move at this speed and radius.
b) If a car is moving at constant speed, the net force acting on the car must be zero. The forces acting on the car are the normal force N, the gravitational force mg, and the force of friction f (since the road is not frictionless).
The vertical force balance equation is still:
N - mg = 0
where N = 8820 N and mg = 900 * 9.8 = 8820 N. Therefore, N and mg cancel each other out.
The horizontal force balance equation is now:
f = F_c = m * v^2 / r
Substituting the values, we get:
f = 900 * 40.0^2 / 100
f = 144000 N
Therefore, the friction force experienced by the car is 144000 N.
F_c = m * v^2 / r
where m = mass of the car = 900 kg, v = speed of the car = 40.0 m/s, and r = radius of the curve = 100 m.
Substituting the values, we get:
F_c = 900 * 40.0^2 / 100
F_c = 144000 N
Since the normal force N is perpendicular to the horizontal surface, we can use trigonometry to find the angle θ.
The vertical force balance equation is:
N - mg = 0
where g = acceleration due to gravity = 9.8 m/s^2. Rearranging, we get:
N = mg = 900 * 9.8 = 8820 N
The horizontal force balance equation is:
F_c = N * sinθ
Substituting the values, we get:
144000 = 8820 * sinθ
sinθ = 16.33
However, this is not a valid answer since we cannot have a sine greater than 1. This means that it is not possible to have a frictionless road that allows the car to move at this speed and radius.
b) If a car is moving at constant speed, the net force acting on the car must be zero. The forces acting on the car are the normal force N, the gravitational force mg, and the force of friction f (since the road is not frictionless).
The vertical force balance equation is still:
N - mg = 0
where N = 8820 N and mg = 900 * 9.8 = 8820 N. Therefore, N and mg cancel each other out.
The horizontal force balance equation is now:
f = F_c = m * v^2 / r
Substituting the values, we get:
f = 900 * 40.0^2 / 100
f = 144000 N
Therefore, the friction force experienced by the car is 144000 N.
A girl threw a ball due west against the wall which is 50.4cm from her position, the
ball bounces due east for 20m.caculate the displacement and distance was covered of the
ball bounces due east for 20m.caculate the displacement and distance was covered of the
ball from the girl to the wall and back to the final position.
The displacement of the ball from the girl to the wall is west (negative direction) and is equal to the distance between the girl and the wall which is 50.4 cm = 0.504 m.
When the ball bounces back it covers a distance of 20 m in the opposite direction (east, positive direction). Therefore, the total displacement of the ball from the girl to the wall and back to the final position is:
displacement = (-0.504) + 20 = 19.496 m
The distance covered by the ball is the total length of its path, which is the sum of the distance from the girl to the wall (50.4 cm) and the distance from the wall to the final position (20 m).
distance = 0.504 + 20 = 20.504 m
The displacement of the ball from the girl to the wall is west (negative direction) and is equal to the distance between the girl and the wall which is 50.4 cm = 0.504 m.
When the ball bounces back it covers a distance of 20 m in the opposite direction (east, positive direction). Therefore, the total displacement of the ball from the girl to the wall and back to the final position is:
displacement = (-0.504) + 20 = 19.496 m
The distance covered by the ball is the total length of its path, which is the sum of the distance from the girl to the wall (50.4 cm) and the distance from the wall to the final position (20 m).
distance = 0.504 + 20 = 20.504 m
1 answer