Answer
given,
initial velocity of skateboard = 5.1 m/s
angle above the horizontal = 55°
height of the ramp = 1 m
a) maximum height of projectile
H = 0.889 m
the maximum height of the skateboard above the ground
= 1 + 0.889
= 1.889 m
b) time to reach the height
t = 0.426 s
horizontal distance = u cos θ × t
= 5.1 × cos 55° × 0.426
horizontal distance = 1.25 m
Answer:
a) 6.95 m/s
b) 1.42 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a) The vertical speed when it leaves the ground. is 6.95 m/s
Time taken to reach the maximum height is 0.71 seconds
Time taken to reach the ground from the maximum height is 0.71 seconds
b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds