Answer:
The velocity of the ball is 3.52 m/s.
Explanation:
A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.
Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;
u = 
Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/
, then;
u = 
= 
u = 3.5185
The velocity of the ball is 3.52 m/s.
Answer:

Explanation:
When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components


now we know that


also we know that


now plug in all data in above equations




Answer: (a) t = 5.44 sec
(b) vf = 53.31 m/s
(c) s = 5.0m
Explanation: from the question, given data
the Height of the tower, h = 145m
from question
(a)
the initial velocity, v₁ = 0 m/s
s = v₁t + 1/2 gt²
-145 m = 0(t) + 1/2 (-9.8t²)
t² = 145/4.9
t² = 29.59
t = 5.44 sec
(b)
the speed of the sphere at the bottom of the tower is
vf² = vi² +2as
vf² = 0 + 2(-9.8 × -145)
vf² = 2842
vf = 53.31 m/s
(c)
when caught, the sphere experiences a deceleration of;
a = -29.0g
the time it would take to decelerate becomes;
vf = vi + at
0 = (53.31) + (-29 ×9.8)t
where t = 53.31 / 284.2
t = 0.1876 sec
∴ the distance travelled during the deceleration becomes;
vf² = vi² + 2as
s = (vf² - vi²) / 2a
s = (0 - 53.31²) / 2×-29×9.8
s = -2841.9561 / -568.4
s = 4.99 ≈ 5.0m
i hope this helps, cheers
Answer:
200N
Explanation:
EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.