Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
The correct answer is d Both the observer's are correct
Explanation:
We know by postulates of relativity that laws of physics are same in different inertial frames.
Thus for each of the frames they make observations related to their frames and since the observations are true for their individual frames they both are correct. But when we compare the two frames we need to use transformation equations to compare both the results.
No work is done because the object needs to be moved. The formula for work is Work = Force x Distance.
