This looks more like projectile motion. In this case, your equation: 5.3m=-0.5(9.8)t^2+15(sin 42°)Δt is not necessary.
It seems to be more useful as Y=Y_{o}+V_{oy}t-\frac{gt^{2}}{2}.
Projectile Motion without Initial Velocity (Only time and angle given)
Homework Statement
A projectile is launched at an angle of 50° above the horizontal and it hits the ground in 12 seconds.
a. Calculate the maximum height of the projectile.
b. Find the range of the projectile...
So, I can just use any velocity?
Or must I use a velocity that intersects?
Considering time is neither given nor required, can I use
V_f = V_i+at?
V_f=0 to find the maximum.
I use the gravities as the slope. However, the lines intersect at the origin, where time and velocity equal...
Homework Statement
A tennis ball is thrown upward with and reaches height of 18 m. What maximum height can reach this ball on the where acceleration of free fall is 6 times less than g? In both cases initial velocity is same. [sic] (excuse my physics teacher's grammar, don't shoot the...
I got it in class. :frown:
This was the process:
(Vf-Vi)/(Tf-Ti)=A
Or rather... ΔV/ΔT=A
Since A= 0.5 m/s2, A=(0.5 m/s)/1 s
Thus, V= 0.5 m/s
The ratio is the same and thus (0.5 m)/(1 s)=(60 m)/(x s)
60 m*s = 0.5x m*s
120 s
:P
Thank you for putting up with me and with...
Homework Statement
A car starts to run from rest with acceleration 0.5 m/s2. How long does it take to travel distance 60m? [sic] (I have a Russian teacher for AP Physics)
Homework Equations
V= (Xf-Xi)/(Tf-Ti)
ΔX= Vi*T+(1/2)aT2
The Attempt at a Solution
I tried to find velocity, but that...
Homework Statement
Solve for ∅
1.33 sin 25.0° = 1.50 sin ∅
Homework Equations
Law of Sines?
The Attempt at a Solution
I did this:
(1.33 sin 25.0°)/1.33 = (1.50 sin ∅)/1.33
sin 25.0° = (1.50 sin ∅)/1.33
but I'm solving for ∅, so I modified a little:
(1.33 sin 25.0°)/1.50 = (1.50 sin...