i got a projectile problem again (1 Viewer)

[wolf7]

a ball is thrown with a velocity of 15m/s at 30 degrees below the horizontal from the top of a 10m building. find:

the time taken to rach the ground

the point where a person must stand to catch the ball without moving

also

a boy standing on the tray of a truck travelling at 10km/h throws a ball vertically upwards. if he catches it again at the same level but 4m horizontal away, how fast was it thrown upwards initally

 

[richz]

hi wolf7,

first one resolve forces, so

u_y=15sin30=7.5m/s
u_x=15cos30=13m/s

time of flight
now use s=u_yt+.5at²
.'. -10=7.5t-4.9t²
rearange and use quadratic formula
so t= 2.39s only, because other answer is negative and obviously u can get a negative time

the point where a person must stand to catch the ball without moving

X=u_xt
=2.39*13
=31m away

a boy standing on the tray of a truck travelling at 10km/h throws a ball vertically upwards. if he catches it again at the same level but 4m horizontal away, how fast was it thrown upwards initally

i would try this, not so sure though.

convert 10km/h to m/s = 2.8m/s
so total time = 4m/2.8m/s
= 1.44s

s=u_yt+0.5at²
.'. 0=u_y1.44-4.9*1.44
so u_y=4.9m/s (not so sure)

 

[helper]

Wolf, have you organised a unifrom way of approaching these questions?
1. Split initial into vertical and horizontal.
2. write down information for x, u_x, v_x, a_x, t
3. write down information for y, u_y, v_y, a_y, t
4. Write down equations for x and y.
5. Look for equation with one unknown to solve or 2 equations you can solve simultaneously.

Common things to remember v_y=0 at maximum height.
a_x=0
x, y, v_x, v_y will change between parts if you are looking at different points along the trajectory.

 

[jono_z1]

first one resolve forces, so

u_y=15sin30=7.5m/s
u_x=15cos30=13m/s

time of flight
now use s=u_yt+.5at²
.'. -10=7.5t-4.9t²
rearange and use quadratic formula
so t= 2.39s only, because other answer is negative and obviously u can get a negative time

You assumed that upwards was positive, since you said displacement = -10m and acceleration due to gravity is -9.8ms^-2. But then wouldn't the downwards (i.e. vertical) velocity have to be negative as well? (i.e. -7.5ms^-1)

then
-10= -7.5t-4.9t²
4.9t²+7.5t-10=0
Using the quadratic forumla,
t=0.86s (the other solution is negative)