I’m sorry if i took up a lot of space, hope this is a valid approximate answer
initial speed of the stuntman is given as

angle of inclination is given as

now the components of the velocity is given as


here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.
So the displacement in vertical direction is given as



by solving above equation we have

Now in the above interval of time the horizontal distance moved by it is given by


since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.
Hard question thx for the points give me brainlest points plz
Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring

Energy lost due to friction

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy


v = 1.40 m/sec