Answer:Answer:
300 seconds
Explanation:
We solve this problem using the equation of uniform motion;
d =v*t
d: Distance traveled
t: Time
v: Speed
We raise the equations of uniform motion because they run at a constant speed.
dJ=3.8*tJ Motion equation for Jenny
dA=4*tA Motion equation for Alyssa
d:distancia
When Jenny and Alissa meet, they will have traveled the same distance,then:
dJ=dA
3.8 tJ =4*tA
tA = * tJ
tA =0.95¨*tJ Equation (1)
Jenny runs 15 seconds longer than Alyssa,then:
tJ = tA+15 Equation (2)
We replace tA =0.95tJ of the Equation (1) in the Equation (2):
tJ = 0.95*tJ +15
tJ-0.95*tJ=15
0.05* tJ=15
tJ=
tJ=300 s
Alyssa catch up with Jenny in 300 s after Jenny’s start .
Given :
Initial velocity, u = 12.5 m/s.
Height of camera, h = 64.3 m.
Acceleration due to gravity, g = 9.8 m/s².
To Find :
How long does it take the camera to reach the ground.
Solution :
By equation of motion :
Putting all given values, we get :
t = 2.56 and t = −5.116.
Since, time cannot be negative.
t = 2.56 s.
Therefore, time taken is 2.56 s.
Hence, this is the required solution.