A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and high
1 answer:
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Answer:
1.6 m/s^2
Explanation:
Hello!
To calculate the acceleration we must know the electric field. The electric field and the potential are related by:
If the particle starts at 2.3m, the electric field is:
E = 36.869 V/m = 36.869 N/C
So, the force on the particle is:
F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N
And its acceleration is :
a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2
Rounded to two significant figures:
1.6 m/s^2
Answer:
The altitude of the plane is 379.5 m.
Explanation:
Initial horizontal velocity, u = 59.1 m/s
Horizontal distance, d = 521 m
let the time taken by the packet to cover the distance is t.
Horizontal distance = horizontal velocity x time
521 = 59.1 x t
t = 8.8 s
let the vertical height is h .
Use second equation of motion in vertical direction.