Answer:
about 42.35 m/s
Explanation:
Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:
which for our case would reduce to:
then the velocity just before hitting would be about 42.35 m/s
How do you measure the wavelength of wave?
2 answers:
You might be interested in
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:
Explanation:
As we know that the formula of range is given as
now we know that
maximum value of the range of the projectile is given as
now we need to find such angles for which the range is half the maximum value
so we will have