A stone is thrown from ground level at 78 m/s. Its speed when it reaches its hig

Question

A stone is thrown from ground level at 78 m/s.
Its speed when it reaches its highest point is
49 m/s.
Find the angle, above the horizontal, of the
stone’s initial velocity


A cannon sends a projectile towards a target
a distance 1330 m away. The initial velocity
makes an angle 25?with the horizontal. The target is hit.
The acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the initial velocity?
Answer in units of m/s
How high is the highest point of the trajectory?
Answer in units of m
How long does it take for the projectile to
reach the target? (Assume no friction)
Answer in units of s

Explanation / Answer

let the angle be x 78cosx = 49 cos x = 0.628 x = 51.08 degree 1330=u*cos25*t initial vertical velocity=u*sin25 vertical acceleration=g=9.8 at highest point the projectile's vertical velocity=0 h=height of the highest point t/2 is the time taken to reach the highest point v=u-g*t 0=u*sin25-9.8*(t/2) u*sin25=4.9*t (u*sin25)/(u*cos25*t)=(4.9*t)/1350 1330*tan25=4.9*t^2 t=11.25 u*sin25=4.9*11.25 u=130.44 m/s v2=u2-2*g*h 0=(130.44*sin25)^2 - 2*9.8*h h=155.05 m highest point of the trajectory=155.05 m

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