Question #227705

A projectile is launched with an initial speed of 46.0 m/s at an angle of 35.0° above the horizontal. The projectile lands on a hillside 3.95 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(a) What is the projectile's velocity at the highest point of its trajectory?


1
Expert's answer
2021-08-19T14:23:13-0400

v0=46msv_0 = 46\frac{m}{s}

α=35°\alpha = 35\degree

t1=3.95st_1 = 3.95s

g=9.8ms2g= 9.8\frac{m}{s^2}

Projectile rise time:\text {Projectile rise time:}

tp=v0sinαg=46sin35°9.8=2,69st_p = \frac{v_0\sin\alpha}{g}= \frac{46*\sin35\degree}{9.8}=2,69s

tp<t1 that is, the projectile reaches the top point of the trajectoryt_p<t_1\text{ that is, the projectile reaches the top point of the trajectory}

 without colliding with the mountainside\text{{ without colliding with the mountainside}}

for tp:\text{for }t_p:

vy=0;vx=v0cosα;v_y = 0;v_x= v_0\cos\alpha;

v=46cos35°=37.68msv = 46*\cos35\degree=37.68\frac{m}{s}


Answer: v=37.68ms\text {Answer: }v =37.68\frac{m}{s}


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