The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s
<h3>
Motion Under Gravity</h3>
The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.
Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.
a. how much later does the ball hit the ground?
The time can be calculated by considering the vertical component of the motion with the use of formula below.
h = ut + 1/2gt²
Where
- Initial velocity u = 0 ( vertical velocity )
- Acceleration due to gravity g = 9.8 m/s²
Substitute all the parameters into the formula
75 = 0 + 1/2 × 9.8 × t²
75 = 4.9t²
t² = 75/4.9
t² = 15.30
t = √15.3
t = 3.9 s
b. how far from the building will it land?
The range can be found by using the formula
R = ut
Where u = 4.6 m/s ( horizontal velocity )
R = 4.6 × 3.9
R = 18 m
c. what is the velocity of the ball just before it hits the ground?
The final velocity will be
v = u + gt
v = 4.6 + 9.8 × 3.9
v = 4.6 + 38.22
v = 42.82 m/s
Therefore, the answers are 3.9 s, 18 m and 42.82 m/s
Learn more about Vertical motion here: brainly.com/question/24230984
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when it reaches the maximum height, all the energy has now been converted into potential energy.when a ball is thrown straight upto into the air,all its initial kinetic energy converted into gravitational potential energy when it reaches its maximum height
Answer:
The answer is 3.48 seconds
Explanation:
The kinematic equation
y= y0+V0*t+1/2*a*(t*t)
-50=0+(0)t+1/2(-9.8)*(t*t)
t=3.194 seconds
During ribbons ball,
x=x0+ Vt+1/2*a*(t*t)
x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)
x= 47.9157m
So, distance (D) = 100-47.9157= 52.084m
52.084m=0+15(t)+1/2*(0)(t*t)
t=52.084/15=3.472286= 3.48seconds
