With an initial velocity of 9.8 m / sg, an object is thrown upwards from the roof of a 100 m high building. Determine maximum height from the ground, time it takes to reach the ground, and the speed with which it reaches the ground.

104.9 m

5.63 s

45.3 m/s

Explanation:

Given:

y₀ = 100 m

v₀ = 9.8 m/s

a = -9.8 m/s²

Find: y when v = 0 m/s

v² = v₀² + 2a (y − y₀)

(0 m/s)² = (9.8 m/s)² + 2 (-9.8 m/s²) (y − 100 m)

y = 104.9 m

Find: t when y = 0 m

y = y₀ + v₀ t + ½ at²

0 m = 100 m + (9.8 m/s) t + ½ (-9.8 m/s²) t²

0 = 100 + 9.8t − 4.9t²

t = [ -9.8 ± √(9.8² − 4(-4.9)(100)) ] / 2(-4.9)

t = (-9.8 ± 45.3) / -9.8

t = 5.63 s

Find: v when y = 0 m

v² = v₀² + 2a (y − y₀)

v² = (9.8 m/s)² + 2 (-9.8 m/s²) (0 m − 100 m)

v = -45.3 m/s

|v| = 45.3 m/s

2b2t

explanation:

2b2t

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle