Respuesta :
The cliff diver jumps off with a velocity of 7.85m/s at an angle of 86° with respect to the ground.
The height of the cliff = 57m
The horizontal distance from the cliff = 2.15m
The time is taken = 4.3s
The acceleration due to gravity g = 9.81m/s²
Consider the vertical y-axis
- d = 57m
- We know the equation
d = ut +1/2gt²
57 = u (4.3) + 1/2 (9.81) (4.3)²
57 = 4.3u + 90.69
u = -7.836m/s
The y component of initial velocity is in the upwards direction
Consider the horizontal x-axis
- d = 2.15 m
- We know the equation, in case of no acceleration
d = ut
2.15 = u (4.3)
u = 0.5 m/s
The net velocity is the square root of the summation of the squares of the x and y components of velocity
Net velocity = 7.85m/s
For the direction, tan θ = y/x = 15.67
θ= 86°
So the cliff diver jumps off with a velocity of 7.85m/s at an angle of 86° with respect to the ground.
Learn more about projectile motion here:
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