From the top of a tower 40 m, a ball is projected upwards with a speed of 20 m/s at an angle of 30° with the horizontal. if g=10m/s^2, after how long will the ball hit the ground?

Question

Shailendra Kumar Sharma · Answer

The velocity is 20m/s and the ball is thrown from a 40m tower

vertical component of velocity is vsin30 =10m/s

so 2^nd law of newton
s=ut+1/2(at2)
-40 =20t - 5t2
Sp 5t2-20t-40=0
t²-4t-8=0

t= 5.5 or -1.46
so the answer is 5.5 s

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From the top of a tower 40 m, a ball is projected upwards with a speed of 20 m/s at an angle of 30° with the horizontal. if g=10m/s^2, after how long will the ball hit the ground?

From the top of a tower 40 m, a ball is projected upwards with a speed of 20 m/s at an angle of 30° with the horizontal. if g=10m/s^2, after how long will the ball hit the ground?

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From the top of a tower 40 m, a ball is projected upwards with a speed of 20 m/s at an angle of 30° with the horizontal. if g=10m/s^2, after how long will the ball hit the ground?

From the top of a tower 40 m, a ball is projected upwards with a speed of 20 m/s at an angle of 30° with the horizontal. if g=10m/s^2, after how long will the ball hit the ground?

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