Question

A ball is dropped from the top of a 94-m-high building. What speed does the ball have in falling 3.8s

Answer 1

Below

Step-by-step explanation:

Quite simply : v = at = 9.81 m/s^2 * 3.8 = 37.3 m/s or 37 m/s ( 2 sig dig)

Answer 2

38m/s

Step-by-step explanation:

We are here given that,

• height of building= 94m
• time = 3.8s
• velocity= ?

since the ball is dropped it will have a zero initial velocity. So we have ,

`\implies u = 0m/s \\`

We may use first equation of motion as ,

`\implies v = u + at \\`

`\implies v = 0 + 10 * 3.8s \\`

`\implies \underline{\underline{ v = 38\ ms^(-1)}} \\`

and we are done!