When 999mm is added to 100m ______ is the result​

a) 19440 km/h²; b) 10 seconds.

We apply the formula S=½gt², where S represents the vertical distance of 74.3m, and g is the acceleration due to gravity valued at 9.8 m/s². Therefore, t=√(2S/g)=3.884 seconds. Consequently, the car remains airborne for around 3.894 seconds. Its horizontal journey is vt=20×3.894=77.88m.

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

The appropriate choice is C.

In physics, the law of gravity helps us understand how gravity varies with height. As altitude increases, so too does the experience of gravity. Changes in altitude also result in variations in weight, though these differences are not particularly significant. Consequently, weighing metals at different heights shows negligible variance as the impact of gravity remains constant across them.