Question

a ball dropped from the top of a building reaches the ground in 3sec calculate -the velocity at which it strikes the building and height of the building plsplspl ans quickly!!!!!!!

Answer 1

Answer:

ғɪʀsᴛ ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ғɪɴᴅ ғɪɴᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴀᴛ ᴡʜɪᴄʜ ɪᴛ sᴛʀɪᴋᴇs ᴛʜᴇ ɢʀᴏᴜɴᴅ :

ᴠ = ᴜ + ɢᴛ

ᴠ = 0 + 10ᴍ/s ^2* 3s

ᴠ = 30ᴍ/s

ʜᴇɪɢʜᴛ :

ʜ = ᴜᴛ + 1/2ɢᴛ^2

ʜ = 0 * 3s + 1/2*10ᴍ/s^2*(3s)^2

ʜ = 1/2*10ᴍ/s^2*9s

ʜ = 45ᴍ

Answer 2

Answer : The height of the building = 45m and velocity of ball when it strikes the ground = 30m/s Step-by-step Explanation : Given : Time ( t ) = 3sec Initial velocity ( u ) = 0 To find : Height of Building (h) = ? Final velocity (v) = ? We know that , S = ut + 1/2 × gt² Substituting the given value in above formula we get, h = 0 × 3 + 1/2 × 10 × (3)² ---------- ( since u = 0 and g = 10m/s² ) = 1/2 × 10 × 9 = 5 × 9 h = 45m Also, v = u + gt Substituting the given value in above formula we get, v = 0 + 10 × 3 = 30 m/s Hence the height of the building and velocity of ball when it strikes the group are 45m and 30m/s respectively .