Answer:
Total time taken=110 seconds
Total distance traveled=480m
Explanation:
First of all, we find the total time taken:
For that, we use the formula : Distance/Speed= Time
Time for part 1 : 200/5=40 seconds
Time for part 2 : 280/4=70seconds
Total time taken=110 seconds
Total distance traveled=480m
Average Speed= 480/110=4.36 m/s
Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)
Average Velocity=-80/110=-0.72 m/s
OR 0.72m/s towards west.
Answer:
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity =
Initial vertical velocity= (Since ball was thrown horizontally only)
Acceleration acting horizontally, = 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, = 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have
Then using Equations of motion for horizontal motion,
Answer:
180.4 m
Explanation:
The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:
h = 100 m
x =?
Height formula h:
Time t is cleared:
t = 4.51 sec
Horizontal distance formula x:
x = 40 m / sec x 4.51 sec
x = 180.4 m
The answer yr looking for would b true!