# Thread: very simply projectile question

1. ## very simply projectile question

1. A particle is projected with an initial velocity of V m/s and attains a maximum height of 12.5m before landing 100m from the point of projection. Calculate the value of V.

Can someone check my answer for this?
2. The displacement, x, of a particle moving in a straight line is represented by the equation: x = 5cos4t
Find a) The initial velocity of the particle
b) initial acceleration of the particle

a) $x = 5cos4t$

$v = x' = -20sin4t$

initial velocity.. when t = 0, v = 0

b) $a = x" = -80cos4t$

when t =0, a = -80

2. Originally Posted by differentiate
1. A particle is projected with an initial velocity of V m/s and attains a maximum height of 12.5m before landing 100m from the point of projection. Calculate the value of V.

use these three equations, all derived from the basic equations for kinematics w/ constant acceleration.

$v_{oy} = \sqrt{2gh}$

$t = \frac{2v_{oy}}{g}$

$v_x = \frac{\Delta x}{t}$

then ...

$v_o = \sqrt{(v_{oy})^2 + (v_x)^2}$

Can someone check my answer for this?
2. The displacement, x, of a particle moving in a straight line is represented by the equation: x = 5cos4t
Find a) The initial velocity of the particle
b) initial acceleration of the particle

a) $x = 5cos4t$

$v = x' = -20sin4t$

initial velocity.. when t = 0, v = 0

b) $a = x" = -80cos4t$

when t =0, a = -80

both correct
...

3. Thanks skeeter. umm... for g, can i say that it's 9.8 m/s? coz someone told me that you have to assume that it's 9.8

4. Originally Posted by differentiate
Thanks skeeter. umm... for g, can i say that it's 9.8 m/s? coz someone told me that you have to assume that it's 9.8
If a numerical answer is required, yes. The question should make that clear.