When the initial speed given is 7.5m/s at an angle of 27° , ball will go
4.637 meters.
Assume no air opposition to the ball ;
Vertical component of ball is sin 27° = 0.453
0.453* 7.5 = 3.404 meters /sec
Time taken to reach ground is :
3.404 = -3.404+9.8*t
t= 6.808/9.8= 0.694 sec
Horizontal component is 7.5*cos27°= 6.682m/s
Distance = speed * time
=6.682 * 0.694
=4.637 meters
Horizontal distance it can cover in 0.694 sec is 4.637 meters
So range of ball is 4.637 meters.
Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.
To know more about projectile motion, refer
brainly.com/question/24216590
#SPJ13
Answer:
The trains mass in pounds would be 40084.029 if you would round it to the hundreths
Explanation:
Answer:
Height.
Explanation:
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
Where,
P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Hence, the property of the object (having a mass of 5 kilograms) which must differ to have different gravitational potential energies is the height from which they are falling from.
The object having the higher height would have a greater gravitational potential energy than the lower object.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
convergent and counterclockwise
hope it helps :)
