Projectile Motion Help (1 Viewer)

luo_ge · Nov 23, 2016

I'm not sure how to find:
• Initial velocity of the projectile
• Initial vertical velocity
• Initial horizontal velocity
• Range
• Maximum height reached
• Time taken to reach the maximum height
• Time of flight
• Final velocity
when only given initial velocity (combined horizontal and vertical) and time

2. A projectile is fired from the top of a 120m high cliff at 25ms-1. It lands on the ground 6.4 sec after firing.

Thanks.

android0 · Nov 24, 2016

$$The initial velocity is given: $25m/s$

$u_y = 25sin(\theta), u_x = 25 sin(\theta)$

$$To decompose into horizontal and vertical components, you need to find $ \theta.$

$$Use $\Delta y = u_yt + \frac{1}{2}a_yt^2$, where $\Delta = -120, u_y = 25sin(\theta), a_y=-9.8, t=6.4$

$\theta = 30.29^\circ$

$u_y = 12.6m/s$

$u_x = 21.6m/s$

$$Use $\Delta x = u_xt $ to find the range. $\Delta x = 138.2m$

$Use $v_y^2 = u_y^2 + 2a_y\Deltay $ to find the maximum height, taking the initial velocity to be 12.6 and the final velocity to be 0 (The turning point of the parabola).$

$\Delta y = 8.1m$

$Maximum height is 128.1m$

$Time to reach maximum height: Use $v_y= u_y + a_yt,$ substitute 0 in $v_y,$ and solve for t.$

$t = 1.29 s$

$Time of flight is given.$

$$Final velocity: Use $v_y= u_y + a_yt,$ substitute 6.4 s into t, and solve for $v_y.$

$v_y = -50.12 m/s$

$x_y = 21.6 m/s$

$Use Pythagoras's formula to find the final velocity.$

$V_f = 54.6 m/s$

$\theta = -66.7^\celc$

Projectile Motion Help (1 Viewer)

luo_ge

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android0

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