i) time of flight
hf=hi+vi'*t-4.9t^2 where hf=hi=0, and vi' is 24.5j
solve for time t.
ii) range= vih*t where vih=10i
d) in the range equation, subtract 20 from the range above, then solve for time t.
hf=vi'*t-4.9t^2 solve for hf
A ball is hit from horizontal ground with velocity (10i+24.5j) ms-1 where the unit vectors i and j are horizontal and vertically upwards respectively.
a)State two assumptions that you should make about the ball in order to make predictions about its motion.
b)the path of the ball is shown in the diagram. (diagram: the ball goes upwards in a curve and then from its highest point downwards in a curve, underneath the curve its straight with a arrow underneath that straight line saying range)
i)show that the time of flight of the ball is 5 seconds
ii)find the range of the ball
c)in fact the ball hits a vertical wall that is 20 metres from the initial position of the ball.
(another diagram in between c and d questions showing the ground and a wall with a curve going from the ground and stopping at the wall, labelled underneath the ground from the starting point of the curve the end point of the wall is an arrow that says range.)
d)Find the height of the ball when it hits the wall.
e)If a heavier ball were projected in the same way, would your answers to part (b) of this question change?Explain why.
Please help, I'm extremely stuck with this whole difficult question and need very clear detailed explanations to help me.
3 answers
air drag negligible
only force i gravity down so horizontal velocity is constant.
Vertical problem:
Vi up = 24.5 m/s
g down = 9.81
then
h = 0 + Vi t - 4.9 t^2
h = 0 when it hits ground
so
24.5 t = 4.9 t^2
t = 0 of course or
t = 24.5/4.9 = 5 seconds in the air
(like they said )
u = constant horizontal velocity = 10 m/s
so
range = 5 seconds * 10 m/s = 50 meters
when is x = 20 meters?
x = 20 = 10 m/s * t
t = 2 seconds
so how high is it?
h = 24.5 *2 - 4.9 (4)
= 49 - 19.6 = 29.4 meters
given our assumptions weight does not matter.
only force i gravity down so horizontal velocity is constant.
Vertical problem:
Vi up = 24.5 m/s
g down = 9.81
then
h = 0 + Vi t - 4.9 t^2
h = 0 when it hits ground
so
24.5 t = 4.9 t^2
t = 0 of course or
t = 24.5/4.9 = 5 seconds in the air
(like they said )
u = constant horizontal velocity = 10 m/s
so
range = 5 seconds * 10 m/s = 50 meters
when is x = 20 meters?
x = 20 = 10 m/s * t
t = 2 seconds
so how high is it?
h = 24.5 *2 - 4.9 (4)
= 49 - 19.6 = 29.4 meters
given our assumptions weight does not matter.
I dont understand:
hf=hi+vi'*t-4.9t^2 where hf=hi=0, and vi' is 24.5j
solve for time t.
ii) range= vih*t where vih=10i
Please write in full clear sentences so i can understand the working out
hf=hi+vi'*t-4.9t^2 where hf=hi=0, and vi' is 24.5j
solve for time t.
ii) range= vih*t where vih=10i
Please write in full clear sentences so i can understand the working out