Ball Throw - uRGENTLy Physics Help Needed?

Question

A ball is thrown with an initial speed of 30 m/s at an angle of 45°.The ball is thrown from a height of 11 m and lands on the ground.

(a) Find the time of flight.
t = s


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(b) Find the maximum height.
h = m


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(c) Find the distance from where the ball is thrown to where it lands.
d = m


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(d) Find the speed at the impact.
v = m/s

Answer 1

The ball will follow
vy(t)=30*sin(45)-g*t
y(t)=11+30*sin(45)-.5*g*t^2

vx(t)=30*cos(45)
x(t)=30*cos(45)*t

a)
when y(t)=0, t=4.8seconds

b) when vy(t)=0, the ball is at apogee
=30*sin(45)-g*t
t=30*sin(45)/g

the height is

y(apogee)=11+.5*30^2*sin^2(45)/g
=34 m

c) Range is
x(4.8)=30*cos(45)*4.8
=102 m

d)
v=sqrt(vx^2+vy^2)
when t=4.8
v=33.4 m/s

j