Respuesta :
Answer:
a)20 m,2s
Explanation:
Given
Initial velocity=20 m /sec
g=10 m/s square
v =0m/s
h =?
t =?
now,
v = U +gt
=v=20 +10t
Or,10t=20
Or,t=20
10
therfore ,2 s.
Again,
h=ut -1 gt square
2
Or, h=20×2-1×10(2)square (multiply and cut )2×5 =10
; 2
Or,40-5×4
therefore , 20m
For question e,
u = 25 ms-1
v = 0 m/s
a = -10m/s^2
We have to find time and distance,
For time,
v = u + at
Which can be rephrased as,
v - u/a = t
t = 0 - 25 / -10
t = 2.5 secs
Hence, the stone takes 2.5 secs to reach the maximum height.
To calculate the maximum height attained by the stone, we have to calculate distance/ displacement .
The formula for calculating distance is,
s = ut + 1/2at^2
s = 25 * 2.5 + 1/2 * -10 * 2.5 * 2.5
s = 62.5 + (- 31.25)
s = 62.5 - 31.25
s = 31.25mts
Hence, the maximum distance travelled by the stone is 31.25 meters.
Hope this helps you. Plz let me know if you have any doubts.
u = 25 ms-1
v = 0 m/s
a = -10m/s^2
We have to find time and distance,
For time,
v = u + at
Which can be rephrased as,
v - u/a = t
t = 0 - 25 / -10
t = 2.5 secs
Hence, the stone takes 2.5 secs to reach the maximum height.
To calculate the maximum height attained by the stone, we have to calculate distance/ displacement .
The formula for calculating distance is,
s = ut + 1/2at^2
s = 25 * 2.5 + 1/2 * -10 * 2.5 * 2.5
s = 62.5 + (- 31.25)
s = 62.5 - 31.25
s = 31.25mts
Hence, the maximum distance travelled by the stone is 31.25 meters.
Hope this helps you. Plz let me know if you have any doubts.