Question
Ben threw a ball up in the air with an initial velocity of 15m/s from a 17m high platform.
A.What is the max height reached before failling down?
B.How long was the ball in the air?
C.What is the final velocity of the ball upon reaching the ground?
A.What is the max height reached before failling down?
B.How long was the ball in the air?
C.What is the final velocity of the ball upon reaching the ground?
Answers
Damon
v =vi - 9.81 t
at top
0 = 15 - 0.981 t
t = 15/9.81 = 1.53 seconds rising
h = 17 + 15(1.53) -4.9(1.53)^2
= 17 + 22.9 - 11.5 = 28.5 meters maxh
how long to fall from there and hoe fast does it hit?
0 = 28.5 - 4.9 t^2
t falling = 5.81 seconds
so totsl time = 5.81+1.53 = 7.34 s
how fast at crash?
v = 0 -9.81(5.81) = - 57 meters/s
at top
0 = 15 - 0.981 t
t = 15/9.81 = 1.53 seconds rising
h = 17 + 15(1.53) -4.9(1.53)^2
= 17 + 22.9 - 11.5 = 28.5 meters maxh
how long to fall from there and hoe fast does it hit?
0 = 28.5 - 4.9 t^2
t falling = 5.81 seconds
so totsl time = 5.81+1.53 = 7.34 s
how fast at crash?
v = 0 -9.81(5.81) = - 57 meters/s
unowen
A. The equation for this would be:
h(t)=-4.9t²+15t+17
The ball will reach its maximum height at -b/2a, where b and a come from the standard equation y=ax²+bx+c. So:
-15/-9.8 is about 1.5 seconds
The height would be :
h=-4.9(1.5)²+15(1.5)+17=28.475 m
B. If h(t)=0, then
-4.9t²+15t+17=0
4.9t²-15t-17=0
using the quadratic formula, we get a positive value for t of 3.94 secs
C. 3.94-1.5=2.44 seconds in free fall.
V=9.8 m/s² x 2.44 seconds=23.912 m/s
☺☺☺☺
h(t)=-4.9t²+15t+17
The ball will reach its maximum height at -b/2a, where b and a come from the standard equation y=ax²+bx+c. So:
-15/-9.8 is about 1.5 seconds
The height would be :
h=-4.9(1.5)²+15(1.5)+17=28.475 m
B. If h(t)=0, then
-4.9t²+15t+17=0
4.9t²-15t-17=0
using the quadratic formula, we get a positive value for t of 3.94 secs
C. 3.94-1.5=2.44 seconds in free fall.
V=9.8 m/s² x 2.44 seconds=23.912 m/s
☺☺☺☺
Blakey
In the time of falling, should it be square rooted? Since there is t^2 left